Sobolev Inequalities and Riemannian Manifolds

University of Connecticut OpenCommons@UConn

Honors Scholar Theses Honors Scholar Program

Spring 5-1-2021

Follow this and additional works at: https://opencommons.uconn.edu/srhonors_theses

Part of the Analysis Commons, and the Geometry and Topology Commons

Recommended Citation Reever, John, "Sobolev Inequalities and Riemannian Manifolds" (2021). Honors Scholar Theses. 824. https://opencommons.uconn.edu/srhonors_theses/824 Sobolev Inequalities and Riemannian Manifolds

Jack Reever

A thesis presented to the Department of Mathematics in partial fulﬁllment of the requirements of a Bachelors of Science with Honors

Storrs, Connecticut Spring 2021 Abstract

Sobolev inequalities, named after Sergei Lvovich Sobolev, relate norms in Sobolev spaces and give insight to how Sobolev spaces are embedded within each other. This thesis begins with an overview of Lebesgue and Sobolev spaces, leading into an introduction to Sobolev inequalities. Soon thereafter, we consider the behavior of Sobolev inequalities on Riemannian manifolds. We discuss how Sobolev inequalities are used to construct isoperimetric inequalities and bound volume growth, and how Sobolev inequalities imply families of other Sobolev inequalities. We then delve into the usefulness of Sobolev inequalities in determining the geometry of a manifold, such as how they can be used to bound a manifold’s number of ends. We conclude with examples of real-world applications Sobolev inequalities.

1 Introduction to Sobolev Spaces

We will begin by introducing Sobolev spaces, and to do so we start with Lebesgue spaces. Let p ∈ [1, ∞). The space of functions which are Lebesgue integrable on a set Ω p-many times is denoted Z Lp(Ω) = {f : |f|p(x)dx < ∞}. Ω These Lebesgue spaces have the norm Z 1/p p ||f||Lp(Ω) = |f| (x)dx . Ω If p = ∞, the space L∞(Ω) is the space of all functions such that |f(x)| < ∞ “almost everywhere” on Ω, which is to say that |f(x)| can be infinite on only a set of measure zero. Here, the norm is the essential supremum of |f(x)| for x ∈ Ω [3]. The elements of a Lebesgue space are equivalence classes, and functions are considered equivalent if they differ only on a set of Lebesgue measure zero. Lebesgue spaces are complete normed spaces. Let k ∈ N ∪ {0} and p ∈ [1, ∞]. The Sobolev Space W k,p(Ω) is defined by

W k,p(Ω): = {u ∈ Lp(Ω) : Dαu ∈ Lp(Ω) ∀α with |α| ≤ k}.

1 This space has the norm X α ||u||W k,p(Ω) := ||D u||Lp(Ω) |α|≤k as discussed in [3]. For a function u to be included in a Sobolev space, all derivatives Dαu must exist and must be represented by an element of Lp(Ω). Note that if k = 0, the space W 0,p(Ω) = Lp(Ω) identically, as the restriction on the derivatives of each element is gone. Sobolev spaces are vector spaces, as elements can be added together and multiplied by scalars, with the resultant functions being elements of the Sobolev space. In addition, the k,p ∞ Sobolev space W0 (Ω) is the completion of C0 (Ω) (the space of infinitely differentiable functions with compact support) with respect to the norm of W k,p(Ω) [3]. Equipped with the definition of Sobolev spaces, we can intuitively embed Sobolev spaces in one another: let Ω ⊂ Rd be a domain with p ∈ [1, ∞) and k ≤ m. Then, W m,p(Ω) ⊂ W k,p(Ω). This is clear because if a function has m derivatives represented in Lp(Ω), then it must have k derivatives in Lp(Ω). Alternatively, for k ≥ 0 and p, q ∈ [1, ∞] with q > p it is clear that W k,q(Ω) ⊂ W k,p(Ω): if we can integrate q−many times, we can integrate p−many times. Now, we will consider the Sobolev Embedding Theorem. Consider W k,p(Rn) with k ∈ N+, p ∈ [1, ∞). If we have two real numbers l, q such that k > l, 1 ≤ p < q < ∞, and 1 k 1 l − = − , p n q n

then W k,p(Rn) ⊆ W l,q(Rn), and this embedding is continuous.

2 2 Introduction to Sobolev Inequalities

When we are working with smooth functions f : R → R with compact support, we have the inequality 1 Z ∞ |f(x)| ≤ |f 0(t)|dt. 2 −∞ This inequality bounds the value of |f(x)| using its derivative. As we move up to multidi- mensional spaces, the notions of absolute value and the derivative become more complicated, and we cannot use this inequality to bound our functions. This brings us to the topic of Sobolev inequalities: how can we construct an inequality that bounds the norm of a function using partial derivatives? While working in Rn, let p ∈ [1, n). The Sobolev conjugate of p is np p∗ := ∈ (p, ∞). n − p

Theorem 2.1: When n ≥ 2, the Sobolev inequality in Rn states Z 1/p∗ Z 1/p |f(x)|p∗ dx ≤ C(n, p) |∇f(x)|pdx (1) Rn Rn for all smooth functions f with compact support, for any p ∈ [1, n). Note that C(n, p) is a constant that depends on n and p [1].

∗ p n Note that this inequality can be written as ||f||Lp (Rn) ≤ C||∇f||L (R ), and moving

forward we will notate these norms as ||f||p∗ ≤ C||∇f||p. To prove Theorem 2.1, we will illustrate Gagliardo and Nirenberg’s proof of the Sobolev inequality in Rn for p = 1, and then expand upon that proof to show this inequality holds n for all real p ∈ [1, n). Note that µ = µn denotes the Lebesgue measure in R .

Proposition 2.2: When n ≥ 2, Z 1/1∗ Z 1/1 |f(x)|1∗ dx ≤ C(n, 1) |∇f(x)|1dx Rn Rn for all smooth functions f with compact support. Note that C(n, 1) is a constant that depends on n.

0 1 1 p p0 Proof. Let 1 ≤ p, p ≤ ∞ with 1 = p + p0 . For f ∈ L (µ), g ∈ L (µ), H¨older’sinequality states that for positive measure µ, Z

fgdµ ≤ ||f||p||g||p0 .

3 pi 1 1 1 By induction, for fi ∈ L , 1 ≤ i ≤ k, 1 ≤ pi ≤ ∞, + + ... + = 1, we see p1 p2 pk Z k Y f1f2...fkdµ ≤ ||fi||pi . (2)

i=1 ∞ n Fix f ∈ C0 (R ): a smooth function with compact support that. By the inequality 1 Z ∞ |f(x)| ≤ |f 0(t)|dt, 2 −∞ for any x = (x1, ..., xn) and any integer i ∈ [1, n] we observe 1 Z ∞ |f(x)| ≤ |∂xi f(x1, x2, ..., xi−1, t, xi+1, ..., xn)|dt. (3) 2 −∞ Fix Z ∞

Fi(x) = |∂xi f(x1, x2, ..., xi−1, t, xi+1, ..., xn)|dt, −∞ which depends on n − 1 coordinates of x with the integration variable replacing the ith coordinate. Additionally, ﬁx

R ∞ R ∞  −∞ ... −∞ |∂xi f(x)|dx1...dxm when i ≤ m Fi,m(x) = R ∞ R ∞  −∞ ... −∞ Fi(x)dx1...dxm when i > m which depends on n − m or n − m − 1 variables, for all integers m ∈ [1, n]. Now, by (3) we have the inequality 1 |f|n ≤ (F ...F ). 2n 1 n Raising both sides of this equality to the power of 1/(n − 1), we see 1 |f|n/(n−1) ≤ (F ...F )1/(n−1). 2n/(n−1) 1 n 1 1 1 1 Setting k = n − 1, p1 = p2 = ... = pn−1 = n − 1, we see + + ... + = (n − 1) = 1. p1 p2 pk n−1 By (2) with induction on m ≤ n we see Z Z 1 ... |f(x)|n/(n−1)dx ...dx ≤ (F (x)...F (x))1/(n−1). 1 m 2n/(n−1) 1,m n,m When n = m, we see n 1/n Y ||f||n/(n−1) ≤ (1/2) ||∂xi f||1 . i=1 Note that both sides were raised to the power of (n − 1)/n when taking the norm in √ n/(n−1) n Qn 1/n 1 Pn Pn L (R ). Because ( 1 ai) ≤ n 1 ai for ai > 0 and n ∈ Z, and 1 |∂xi f| ≤ n|∇f|, this becomes n 1 X 1 ||f|| ≤ ||∂ f(x)|| dx ≤ √ ||∇f|| . n/(n−1) 2n xi 1 2 n 1 i=1 This proves Proposition 2.2, the Sobolev Inequality in Rn for p = 1. [1]

4 Next, we are going to build oﬀ of the above proof to show that Theorem 2.1 (1) holds for p ≥ 1.

Proposition 2.3: If Theorem 2.1, the Sobolev Inequality in Rn, holds for p = 1, then it holds for all p ∈ [1, n).

Proof. Knowing that (1) holds for p = 1, or that 1 ∀f ∈ C∞( n) ||f|| ≤ √ ||∇f|| , (4) 0 R n/(n−1) 2 n 1

∞ n α 1 ﬁx a p > 1. For any α > 1 and function f ∈ C0 (R ), |f| is C0 and satisﬁes

|∇|f|α| = α|f|α−1|∇f|.

α The function |f| can be approximated by a sequence of smooth functions (fi) ∈ C0 such α α that ∇fi → ∇|f| . This allows us to substitute |f| into (4): 1 Z ||f||α ≤ √ α |f(x)α−1|∇f(x)|dx αn/(n−1) 2 n 1/p0 1/p 1 Z 0 Z ≤ √ α |f(x)|(α−1)p dx |∇f(x)|pdx 2 n

0 (n−1)p with 1/p + 1/p = 1. Setting α = (n−p) implies

1 (n − 1)p ||f||(n−1)p/(n−p) ≤ √ ||f||n(p−1)/(n−p)||∇f|| . np/(n−p) 2 n n − p np/(n−p) p

n(p−1)/(n−p) To simplify, divide both sides by ||f||np/(n−p) . We see that

(n − 1)p n(p − 1) pn − p − np + n n − p − = = = 1, n − p n − p n − p n − p and so 1 (n − 1)p ||f|| ≤ √ ||∇f|| . p∗ 2 n n − p p We have thereby proven a Sobolev inequality in Rn for p > 1: for any integer n ≥ 2 and p ∈ [1, n), then ∞ n np − p ∀f ∈ C ( ) ||f|| ∗ ≤ √ ||∇f|| . 0 R p 2(n − p) n p

5 We have shown the Sobolev Inequality in Rn holds for p ∈ [1, n), and thus have proven Theorem 2.1 [1]. Alternatively, in the case that n < p ≤ ∞, we have the H¨oldercontinuity estimate: |f(x) − f(y)| Z 1/p sup ≤ C |∇f(x)|pdx n 1−n/p x,y∈R |x − y| Rn for smooth functions f with compact support, for some constant C that depends only on n and p. We can generalize Sobolev inequalities to subsets of Rn: let’s consider the general n n Sobolev inequality for k < p . Let Ω ⊂ R be Lipschitz, with k ∈ {1, ..., n − 1} and n p ∈ [1, k ). Let 1 1 k np = − =⇒ q = > p. q p n n − kp Then, W k,p(Ω) ⊂ Lq(Ω) and

||u||Lq(Ω) ≤ C||u||W k,p(Ω) for some constant C that depends on k, p, n, and Ω. It is clear that Lq ⊂ Lp, but this inequality tells us that W k,p ⊂ Lq ⊂ Lp. The Gagliardo-Nirenberg-Sobolev inequality states for p ∈ [1, n),

p(n − 1) ||u|| p∗ n ≤ ||Du|| p n L (R ) n − p L (R )

1 n for u ∈ C0 (R ). Various Sobolev inequalities are related to one another; for example, there are equivalences between “weak” and “strong” Sobolev inequalities. Consider the Nash inequality on a Riemannian manifold M:

∞ 1+2/v 2/v ∀f ∈ C0 (M), ||f||2 ≤ C||∇f||2||f||1 this inequality is equivalent to the Sobolev inequality

∞ ∀f ∈ C0 (M), ||f||2v/(v−2) ≤ C||∇f||2.

We will examine more examples of inequality equivalences in further detail in Section 3.

6 3 Sobolev Inequalities on Manifolds

Sobolev inequalities were prevalent during the development of analysis on manifolds because Fourier analysis and other tools in Euclidean space are no longer available on manifolds. On manifolds, the original question of which inequalities are true or not is expanded into a search for necessary and suﬃcient conditions for Sobolev inequalities to hold true. Not every manifold admits global Sobolev inequalities; a workaround is to use families of local Sobolev inequalities. For example, we can construct a Sobolev inequality locally on a ball; for a ball B = B(x, r) on a complete Riemannian manifold, there is a constant C(B) such that for smooth functions f with compact support in B

Z 2/q 2 Z q C(B)r 2 −2 2 |f| dµ ≤ 2/v (|∇f| + r |f| )du B µ(B) B 1 1 1 for constants q, v > 2 such that q = 2 − v .

3.1 Isoperimetric Inequalities

Geometric properties of manifolds can be determined using isoperimetric inequalities, inequalities that relate the measure of a set’s boundary to the set’s measure or volume. For some background, consider R2. A circle is the 2-dimensional or planar shape that maximizes area (A) in a ﬁxed perimeter (circumference C). These two values are related by the equation

4πA = C2 derived from C = 2πr, A = πr2. Thus, the isoperimetric inequality in R2 tells us that for a closed curve of length L, with A the area of the region enclosed by L,

L2 ≥ 4πA.

Let Ω be a bounded subset of a Riemannian manifold M, with a smooth boundary ∂Ω. We want to find the maximal volume that can be enclosed in a hypersurface of a fixed (n − 1)-measure. Using the same intuition as the R2 example, in Euclidean space an n−dimensional sphere or ball maximizes enclosed volume with a fixed surface area. So, if M is n−dimensional Euclidean space, we can derive an isoperimetric inequality from the equation that defines the volume of a sphere in n dimensions. Now we will consider the isoperimetric inequality for a Riemannian manifold that may not be in Euclidean space. Let M be an n−dimensional Riemannian manifold. If M satisfies

1−1/v µn(Ω) ≤ C(M, v)µn−1(∂Ω) (5)

7 for any Ω ⊂ M with a smooth boundary for some v ≥ n, C(M, v) > 0, then rv V (x, r) ≥ (6) (vC(M, v))v

with V (x, r) denoting the volume of a ball with radius r [1]. This is because ∂rV (x, r) is the (n − 1)−dimensional Riemannian volume of the ball B(x, r)’s boundary: in general, the volume of a sphere in n dimensions is

bn/2c n V (x, r) = Cnπ r for some constant Cn [2]. Meanwhile, this sphere’s “surface area” can be found by the formula bn/2c n−1 An = Cnπ nr ,

which is equivalent to ∂r(V (x, r)) [2]. We see that

1−1/v V (x, r) ≤ C(M, v)∂rV (x, r) 1 =⇒ V (x, r)1−1/v ≤ C(M, v)V (x, r)n r 1 =⇒ V (x, r)−1/v ≤ C(M, v)n r r =⇒ V (x, r)1/v ≥ nC(M, v) rv rv =⇒ V (x, r) ≥ ≥ nvC(M, v)v vvC(M, v)v because v ≥ n.

Theorem 3.1: A manifold M satisﬁes the conditions in (5) for some v, C(M, v) > 0 if and only if ∞ ∀f ∈ C0 (M), ||f||v/(v−1) ≤ C(M, v)||∇f||1. This is an (L1, v)-Sobolev inequality [1].

Definition 3.2: If we fix p, v such that 1 ≤ p < v, a Riemannian manifold M satisfies an (Lp, v)-Sobolev inequality if there exists a constant C(M, p, v) that satisfies

∞ ∀f ∈ C0 (M), ||f||pv/(p−v) ≤ C(M, p, v)||∇f||p.

Now, we have a deﬁnition of the (Lp, v)−Sobolev inequality. Let us consider the following theorem, which states that (Lp, v)−Sobolev inequalities imply other inequalities of this type.

8 Theorem 3.3: If M satisﬁes an (Lp, v)-Sobolev inequality, then M satisﬁes (Lq, v)-Sobolev inequalities for all q ∈ [p, v).

Proof. Let γ > 1 and use the (Lp, v) inequality on |f|γ:

Z 1/p γ p(γ−1) p ||f||γpv/(v−p) ≤ γC(M, p, v) |f| |∇f| dµ . M By the H¨olderinequality, we see Z p(γ−1) p p(γ−1) p |f| |∇f| ≤ |||f| ||q/(q−p)|||∇f| ||q/p

which implies Z 1/p Z 1/p−1/q Z 1/q |f|p(γ−1)|∇f|p dµ ≤ |f|pq(γ−1)/(q−p)dµ |∇f|qdµ . M M M Using these two inequalities, we see Z 1/p−1/q Z 1/q γ pq(γ−1)/(q−p) q ||f||γpv/(v−p) ≤ γC(M, p, v) |f| dµ |∇f| dµ M M

q(n−p) and if we select γ = p(n−q) we ﬁnd γ − 1 = n(q − p)/p(n − q), implying q(n − p) ||f|| ∗ ≤ C(M, p, v)||∇f|| , q p(n − q) q an (Lq, v)-Sobolev inequality [1].

3.2 Bounding Volume Growth with Sobolev Inequalities

Let M be a manifold that satisﬁes an (Lp, v)-Sobolev Inequality:

∞ ∀f ∈ C0 (M), ||f||pv/(v−p) ≤ C(M, p, v)||∇f||p.

1 1 1 Let q = vp/(v − p) =⇒ q = p − v , and let r, s ∈ (0, ∞) and θ ∈ [0, 1] such that 1 1 1 1 − θ θ 1 − θ = θ − + = + . r p v s q s The H¨olderinequality tells us θ 1−θ ||f||r ≤ ||f||q||f||s and because q = vp/(v − p) we apply the (Lp, v)-Sobolev inequality above to ﬁnd

θ 1−θ ||f||r ≤ (C(M, p, v)||∇f||p) ||f||s .

9 Thus, if M satisﬁes an (Lp, v)-Sobolev inequality, it also satisﬁes the above inequality for all 1 1 1 1−θ r, s ∈ (0, ∞) and θ ∈ [0, 1] such that r = θ p − v + s [1]. We want to use this result to generalize the isoperimetric inequality (6). In the introduction to this section, we discussed that not all manifolds admit global Sobolev inequalities, and at times we have to use local inequalities based in balls. Now, we will use the above result to place a lower bound on the volume of a ball B(x, r). We will notate the volume growth function as V (x, r) = µ(B(x, r)), the measure of the ball of radius r centered at x ∈ M. The following result provides a lower bound for the volume growth V (x, t) in M, as seen in [1, Theorem 3.1.5].

Theorem 3.4: Assume that on a manifold M

∞ θ 1−θ ∀f ∈ C0 (M), ||f||r ≤ (C(M, p, v)||∇f||p) ||f||s

is satisfied for some r, s with 0 < s ≤ r ≤ ∞ and some θ ∈ (0, 1], such that θ 1 − θ 1 + − > 0. p s r Then, for v defined by θ θ 1 − θ 1 = + − v p s r we have tv t v V (x, t) ≥ = . 2v2/θr+v/θC(M, p, v)v 2v/θr+1/θC(M, p, v) Particularly, if a manifold M satisfies an (Lp, v)-Sobolev inequality with 1 ≤ p < v, then n ≥ v and inf {tvV (x, t)} > 0. x∈M t>0 Proof. For a fixed x ∈ M and t > 0, consider

f(y) = max{t − d(x, y), 0}.

Then, t ||f|| ≥ V (x, t/2)1/r r 2 1/s ||f||s ≤ tV (x, t) 1/p ||∇f||p ≤ V (x, t) .

10 and thus 1 V (x, t)θ/p+(1−θ)/s ≥ (t/C)θV (x, t/2)1/r. 2 Here, deﬁne v by θ θ 1 − θ 1 = + − v p s r to ﬁnd V (x, t) ≥ (2Cθ)−rv/(v+θr)tθrv/(v+θr)V (x, t/2)v/(v+θr).

For a = v/(v + θr), we ﬁnd

V (x, t) ≥ (2Cθ)−ratθraV (x, t/2)a

θ −r Pi aj θr Pi aj −θr Pi (j−1)aj i ai ≥ (2C ) 1 t 1 2 1 V (x, t/2 ) .

Additionally,

∞ ∞ 2 X a v X a v2 aj = = , (j − 1)aj = = , 1 − a θr 1 − a θ2r2 j=1 j=1 so as we allow i → ∞ we ﬁnd v 2 t V (x, t) ≥ (2Cθ)−v/θtv2−v /(θr) = . 2v/θ+v2/(θr)Cv

3.3 Weak and Strong Sobolev Inequalities

In order to prove that a certain manifold satisfies a Sobolev inequality, at times we must use the equivalences between weak and strong forms of Sobolev-type inequalities that were introduced in Section 2. As demonstrated in subsection 4.1, we know that if M satisfies an (Lp, v)-Sobolev inequality, then M satisfies (Lq, v)-Sobolev inequalities for all q ∈ [p, v). Any Sobolev inequality can be used in conjunction other inequalities, such as the Hölder inequality, to lead to “weaker” inequalities. Moving forward, we will discuss how “weaker” Sobolev inequalities imply stronger Sobolev inequalities. In the previous subsection, we saw the inequality

∞ 1/r θ 1/s 1−θ ∀f ∈ C0 (M), sup{tµ(|f| > t) } ≤ (C(M, p, v)||∇f||p) (||f||∞µ(supp(f)) ) . t>0

∗,θ This is called the (Sr,s) inequality, and is the weakest Sobolev inequality discussed so far. θ The (Sr,s) inequality is stronger and states

∞ θ 1−θ ∀f ∈ C0 (M), ||f||r ≤ (C(M, p, v)||∇f||p) ||f||s .

11 1 Here, 0 < r, s ≤ ∞ and θ ∈ (0, 1]. When θ = 1, we ﬁnd (Sr,s) states

∞ ∀f ∈ C0 (M), ||f||r ≤ C(M, p, v)||∇f||p

∗,1 and (Sr,s ) states

∞ 1/r ∀f ∈ C0 (M), sup{tµ(|f| > t) } ≤ C(M, p, v)||∇f||p. t>0

∗,θ0 We will show how a weak inequality (Sr0,s0 ) implies a collection of stronger inequalities. First, consider the parameter q = q(r, s, θ) 6= 0 that satisﬁes

1 θ 1 − θ = + r q s

1 1 1 Recall that this q was introduced at the beginning of subsection 4.2 and satisfies q = p − v . We ∗,θ0 will prove that any weak inequality (Sr0,s0 ) for fixed r0, s0, θ0 implies all stronger inequalities θ ∗,θ0 (Sr,s) for all r, s ∈ (0, ∞] and θ ∈ (0, 1] with q(r, s, θ) = q(r0, s0, θ0): an inequality (Sr0,s0 ) θ for fixed r0, s0, θ0, and p0 implies all inequalities (Sr,s) for all p ≥ p0 with r, s, θ, p satisfying 1 θ 1−θ 1 1 1 r = q + s and q = p − v .

3.3.1 Equivalences when q ∈ (0, ∞)

∗,θ0 Begin by ﬁxing p ∈ [1, ∞). Assume that, on a manifold M,(Sr0,s0 ) is satisﬁed for some 1 θ 1−θ 0 < r0, s0 ≤ ∞ and θ ∈ (0, 1]. Let q = q(r0, s0, θ0) satisfy r = q + s with p ≤ q < ∞. Let ∞ f ∈ C0 (M) be nonnegative and set

k k fρ,k = (f − ρ )+ ∧ ρ (ρ − 1)

for any ρ > 1 and k ∈ Z. Here, u+ = max{u, 0} and u ∧ v = min{u, v}: fρ,k reads as

min{(max{f − ρk, 0}), ρk(ρ − 1)}.

k k+1 We see that |∇fρ,k| ≤ |∇f|. Because {fρ,k ≥ (ρ − 1)ρ } = {f ≥ ρ }, and because M

∗,θ0 satisﬁes (Sr0,s0 ) we see

k k+1 1/r0 θ0 k k 1/s0 1−θ0 (ρ − 1)ρ µ({f ≥ ρ }) ≤ (C(M, p, v)||∇f||p) )((ρ − 1)ρ µ({f ≥ ρ }) ) .

12 Set N(f) = sup{ρkµ({f ≥ ρk})1/q}. We then ﬁnd k

θ0 C||∇f||p µ({f ≥ ρk+1})1/r0 ≤ ρ−k(θ0+(1−θ0)q/s0) N(f)q(1−θ0)/s0 ρ − 1 θ0 C||∇f||p ≤ ρ−kq/r0 N(f)q(1−θ0)/s0 ρ − 1 θ0 C||∇f||p =⇒ N(f)q/r0 ≤ ρq/r0 N(f)q(1−θ0)/s0 ρ − 1 ρq/r0 =⇒ N(f) ≤ C||∇f|| ρ − 1 p by the deﬁnition of q. This tells us

1+q/r0 1/q ρ sup{tµ({f ≥ t}) } ≤ ρN(f) ≤ C||∇f||p. t>0 ρ − 1

When we set ρ = 1 + r0θ0/q this gives 1/q q sup{tµ({f ≥ t}) } ≤ e 1 + C||∇f||p. (7) t>0 r0θ0

p This is the “weak” form of the (L , v)-Sobolev inequality ||f||pv/(v−p) ≤ AC||∇f||p. We

∗,θ0 θ want to show that if M satisﬁes (Sr0,s0 ), then all inequalities (Sr,s) with 0 < r, s ≤ ∞, θ ∈ (0, 1] with q(r, s, θ) = q are satisﬁed. Particularly, we want to show that there is a constant A such that

∞ ∀f ∈ C0 (M), ||f||q ≤ AC||∇f||p, (8)

p pv an (L , v)-Sobolev inequality (recall q = p−v ). We move forward using the fact that for 1 ≤ p < q < ∞, if

∞ 1/q ∀f ∈ C0 (M), sup{tµ({f ≥ t}) } ≤ C1||∇f||p t>0 then ∞ ∀f ∈ C0 (M), ||f||q ≤ 2(1 + q)C1||∇f||p. Using this with (7) we ﬁnd the desired result (8), with A = 2e(1 + q)1 + q and C r0θ0 ∗,θ0 the constant from (Sr0,s0 ).

3.3.2 Equivalences when q = ∞

1 θ 1−θ s Again, let us ﬁx p ∈ [1, ∞). When q = ∞, the relationship r = q + s becomes r = 1−θ . s ∗,1−s0/r0 Thus, θ = 1 − r . We want to show that if (Sr0,s0 ) is satisﬁed on a manifold M for some

13 1−s/r 0 < s0 < r0 < ∞, then all inequalities (Sr,s ) with 0 < s < r < ∞ are satisﬁed on M. ∞ Following the same approach as the previous sub-subsection, if f ∈ C0 (M) is nonnegative and we consider fρ,k, we ﬁnd

r0−s0 C||∇f||p ρkr0 µ({f ≥ ρk+1}) ≤ ρks0 µ({f ≥ ρk}). ρ − 1

t Set a real number t > 0 and set r = r0 + t, s = s0 + t. Multiplying both sides by ρ brings us to C||∇f|| r0−s0 ρkrµ({f ≥ ρk+1}) ≤ p ρksµ({f ≥ ρk}). ρ − 1

Note that r0 − s0 = r − s because r0 = r − t, s0 = s − t. If we sum over all k, we see r−s X C||∇f||p X ρkrµ({f ≥ ρk+1}) ≤ ρksµ({f ≥ ρk}). ρ − 1 k k Building oﬀ of these sums, we can ﬁnd Z ρk+2 1 r 1 X s−1 kr k+1 2r r ||f||r ≤ 2r r r t µ({f ≥ t})dt ≤ ρ µ({f ≥ ρ }) ρ − ρ ρ − ρ k+1 k ρ and X ρs ρksµ({f ≥ ρk}) ≤ ||f||s. ρs − 1 s k

Putting this together, we ﬁnd for all t > 1, r = r0 + t, s = s0 + t ρr+s(ρr − 1) ||f||r ≤ (C||∇f|| )r−s||f||s r (ρs − 1)(ρ − 1)r−s p s

ρr+s(ρr − 1) 1/r =⇒ ||f|| ≤ (C||∇f|| )1−s/r||f||s/r r (ρs − 1)(ρ − 1)r−s p s 1−s/r which is the deﬁnition of (Sr,s ), with a constant depending on the ﬁxed ρ > 1. Thus, all

1−s/r ∗,1−s0/r0 inequalities (Sr,s ) with s0 ≤ s < r < ∞, r0 ≤ r are satisfied on M if M satisfies (Sr0,s0 ) θ for some 0 < s0 < r0 < ∞. Meanwhile, the Hölderinequality shows that if r ≥ s,(Sr,s) θ0 0 0 0 0 0 ∗,1−s0/r0 implies (Sr0,s0 ) for all r , s such that s ≤ r ≤ r and s ≤ s. Thus, if M satisfies (Sr0,s0 ), 1−s/r M satisfies all (Sr,s ) with 0 < s < r < ∞.

∗,θ0 For all bounded sets Ω ⊂ M and all r ≥ 1, if M satisﬁes (Sr0,s0 ) for some 0 < s0 < r0 < ∞

then there exists a constant A1 such that

∞ 1/r ∀f ∈ C0 (Ω), ||f||r ≤ A1(1 + r)µ(Ω) ||∇f||p.

In addition, there exist constants α > 0 and A2 such that Z ∞ α|f|/||∇f||p ∀f ∈ C0 (Ω), e dµ ≤ A2µ(Ω) Ω

14 3.3.3 Equivalences when q ∈ (−∞, 0)

∗,θ0 Begin by ﬁxing p ∈ [1, ∞). Assume that on a manifold M,(Sr0,s0 ) is satisﬁed for some

0 < s0 < r0 ≤ ∞ and θ0 ∈ (0, 1]. Again, take q = q(r0, s0, θ0) as before. With s0 < r0, we θ have q ∈ (−∞, 0). We want to show that all inequalities (Sr,s) with 0 < s < r ≤ ∞, θ ∈ (0, 1]

and q(r, s, θ) = q(r0, s0, θ0) are satisﬁed: particularly, we want to show that there is a constant A ∈ R such that

∞ 1/(1−s/q) 1/(1−q/s) ∀f ∈ C0 (M), ||f||∞ ≤ A(C||∇f||p) ||f||s for all s ∈ (0, ∞), with C the constant from (S∗,θ0 ). Set a nonnegative function f ∈ C∞(M) f0,s0 0 with ||f||∞ 6= 0. In adition, ﬁx ε > 0 and ρ > 1, and deﬁne k(f) to be the largest integer k k such that ρ < ||f||∞. Then, set

k k−1 fρ,k = min{(max{f − ||f||∞ + ε + ρ , 0}), ρ (ρ − 1)} k k−1 = (f − (||f||∞ − ε − ρ ))+ ∧ ρ (ρ − 1)

k for all k ≤ k(f). Fix λk = ||f||∞ − ε − ρ . Note that fρ,k has support in {f ≥ λk} and

k−1 {fρ,k ≥ ρ (ρ − 1)} = {f ≥ λk−1}.

∗,θ0 Because M satisﬁes (Sr0,s0 ), apply this inequality to fρ,k to ﬁnd

θ0 C||∇f||p ρk−1µ({f ≥ λ })1/r0 ≤ ρ(k−1)(1−θ0)µ({f ≥ λ })(1−θ0)/s0 k−1 ρ − 1 k θ0 C||∇f||p =⇒ µ({f ≥ λ })1/r0 ≤ ρ(1−θ0)µ({f ≥ λ })(1−θ0)/s0 k−1 ρ − 1 k

k−1 qk δ(k−1) by dividing by ρ . Now, set ak = ρ µ({f ≥ λk}) and multiply by ρ with r0(1+δ) = q to ﬁnd θ0 C||∇f||p a1/r0 ≤ ρ−q(1−θ0)/s0 a(1−θ0)/s0 k−1 ρ − 1 k holds for all k ≤ k(f). Note that ak > 0 for all k ≤ k(f), and lim ak = ∞. Set k→−∞ a = inf ak > 0 to ﬁnd k≤k(f)

θ0 C||∇f||p a1/r0 ≤ ρ−q(1−θ0)/s0 a(1−θ0)/s0 ρ − 1 which implies θ0 2 C||∇f||p a ≥ ρ−q (1−θ0)/(s0θ0) ρ − 1

15 s s Because λ µ({f ≥ λ}) ≤ ||f||s, we ﬁnd

θ0 s C||∇f||p λ−sρpk||f||s ≥ ρ−q (1−θ0)/s0θ0 k s ρ − 1

1 Letting ε = 0 and setting k = k(f) − 1, ρ = 1 + 1+|q| we ﬁnd

(1−θ0)/s0θ0 1/(1−s/q) 1/(1−q/s) ||f||∞ ≤ 4(1 + |q|)(e C||∇f||p) ||f||s .

θ Thus, all inequalities (Sr,s) with 0 < s < r ≤ ∞, θ ∈ (0, 1] with q(r, s, θ) = q(r0, s0, θ0) are satisﬁed on M.

∗,θ0 In addition, if M satisﬁes (Sr0,s0 ) with q ∈ (−∞, 0) then there exists a constant A such that ∞ −1/q ∀f ∈ C0 (Ω), ||f||∞ ≤ ACµ(Ω) ||∇f||p

∗,θ0 for bounded domains Ω ∈ M, with C the constant from (Sr0,s0 ).

3.3.4 Equivalences under diﬀerent values of p

In the previous sections when we examined equivalences between weak- and strong-Sobolev ∗,θ inequalities, we ﬁxed p and found equivalences under the same value of q. Allow (Sr,s (p)) θ ∗,θ θ and (Sr,s(p)) refer to the inequalities (Sr,s ) and (Sr,s) as we will now vary p ∈ [1, ∞).

∗,θ0 Assume that on a manifold M, the inequality (Sr0,s0 (p0)) is satisﬁed for some p0 ∈

[1, ∞), 0 < r0, s0 ≤ ∞, 0 ∈ (0, 1]. Assume that q0 = q(r0, s0, θ0) satisfies 1/q0 < 1/p0, and let v = p0q0 1 = 1 − 1 . By the results of the previous sub-subsections, we know the q0−p0 q0 p0 v σ0 inequalities (Sp0,u0 (p0)) are satisfied for all u0, σ0 such that q(p0, u0, σ0) = q0. ∞ σ0 γ Let f ∈ C0 (M) and γ ≥ 1. Apply some (Sp0,u0 (p0)) to |f| to find

Z σ0/p0 γ (γ−1)p0 p0 γ(1−σ0) ||f||p0γ ≤ C γ |f| |∇f| dµ ||f||u0γ . M The H¨olderinequality states

Z 1/p0 (γ−1)p0 p0 γ−1 f g dµ ≤ ||f||p0γ ||g||p0γ,

which brings us to

γ σ0/p0 (γ−1)σ0 σ0 γ(1−σ0) ||f||p0γ ≤ Cγ ||f||p0γ ||∇f||p0γ||f||u0γ

σ0 pu0 σ If we set p = γp0, σ = , u = we see that the inequality (S (p))is satisﬁed. In σ0+γ(1−σ0) p0 p,u addition, we have 1 − σ 1 − σ = γ 0 , σ σ0

16 σ0 σ and that the parameter v has not changed from (Sp0,u0 (p0)) to (Sp,u(p)); 1 1 1 1 1 = − = − . v p0 q0 p q(p, u, σ)

σ0 σ Thus, if M satisfies (Sp0,u0 (p0)), M satisfies all (Sp,u(p)) such that 1 1 1 = − . q(p, u, σ) p v This allows us to increase p and construct a strong Sobolev inequality. In addition, the results from the previous three sections show that even under the new value of p, all inequalities θ 1 1 1 (Sr,s(p)) with 0 < r, s ≤ ∞, θ ∈ (0, 1] such that q(r,s,θ) = p − v are satisfied.

3.4 Pseudo-Poincar´eInequalities

A pseudo-Poincaréinequality is tool that can be used to prove a manifold satisfies a Sobolev inequality. Here, we will provide a brief overview of what a psuedo-Poincaréinequality is, and demonstrate its ability to prove what Sobolev inequalities a manifold satisfies. In the next section, a connection between a manifold’s geometry and a psuedo-Poincaréinequality is illustrated. ∞ For any f ∈ C0 (M), set 1 Z ft(x) = f(y)dµ(y). V (x, t) B(x,t) It is said that M satisfies a psuedo-Poincaréinequality in Lp if there exists a constant C such that ∞ ∀f ∈ C0 (M), ∀t > 0, ||f − ft||p ≤ Ct||∇f||p.

||f||1 Note that with this deﬁnition of ft, we have ||f||∞ ≤ V (x,t) . The following result provides a connection between psuedo-Poincar´einequalities and Sobolev inequalities of the θ (Sr,s(p))−type, as seen in [1, Theorem 3.3.1].

Theorem 3.5: Assume that M satisﬁes

∞ ∀f ∈ C0 (M), ∀t > 0, ||f − ft||p ≤ Ct||∇f||p. for some p0 ∈ [1, ∞) and that inf {t−vV (x, t)} > 0 x∈M t>0 θ for some v > 0. Then the inequalities (Sr,s(p)) are satisﬁed on M for all p ≥ p0, and all 0 < r, s ≤ ∞, θ ∈ (0, 1] such that q(r, s, θ) satisﬁes 1/q = 1/p − 1/v.

17 Proof. By the result of sub-subsection 3.3.4 Equivalences under diﬀerent values of p, we only

have to treat the case p = p0. Let M be a manifold that satisﬁes

∞ ∀f ∈ C0 (M), ∀t > 0, ||f − ft||p ≤ Ct||∇f||p

v for some p0 ∈ [1, ∞). There exists a constant c > 0 such that V (x, t) ≥ (ct) for all ∞ x ∈ M, t > 0. Set a nonnegative f ∈ C0 (M) and a constant λ > 0. For any t > 0, write

µ({f ≥ λ}) ≤ µ({|f − ft| ≥ λ/2}) + µ({ft ≥ λ/2})

||f||1 and set t such that (ct)v = λ/4. Then, µ({ft ≥ λ/2}) = 0 and

µ({f ≥ λ}) ≤ µ({|f − ft| ≥ λ/2}) p p ≤ (2/λ) ||f − ft||p p ≤ (2Ct||∇f||p/λ) 1+2/v 1/v −1−1/vp = (2 C/c)||f||1 ||∇f||pλ .

and so 21+2/vC p λ(vp+p)/vµ({f ≥ λ}) ≤ ||f||1/v||∇f|| . c 1 p v Raising this to the power p+pv we ﬁnd

v 1/(1+v) p+pv v/(1+v) v/(1+v) λµ({f ≥ λ}) ≤ 4(C/c) ||f||1 ||∇f||p ,

∗,v/(1+v) which is the inequality S(pv+p)/v,1(p) . By the results of the previous sub-subsections, this proves Theorem 3.5.

Thus, the psuedo-Poincaréinequality is useful to show what Sobolev inequalities are satisfied on a manifold. In the next section, we will see that there is a geometric sufficient condition to show that a manifold satisfies a psuedo-Poincaréinequality.

18 4 Sobolev Inequalities and Manifold Geometry

In this section, we are going to investigate the information about a Riemannian manifold’s geometry that is encoded in the manifold’s Sobolev-type inequalities. We will begin with some background in manifold geometry. On a complete Riemannian manifold (M, g), you can start at any point p ∈ M and follow a “straignt” line in any direction indefinitely. The notation (M, g) refers to the smooth manifold M and the Riemannian metric g, a positive-definite inner product tensor that provides a notion of distance between two points of M. Manifolds are known as spaces that locally resemble Euclidean space. This conceptual- ization begs the question: how different is my manifold from Euclidean space? To answer this question, the Ricci curvature tensor was developed. The Ricci curvature tensor R provides a measure of how different the geometry of a manifold is from the geometry of ordinary Euclidean space. Lower bounds on the Ricci tensor can provide information on the global geometry and topology of the manifold. The Ricci curvature tensor is a symmetric two-tensor obtained by the contraction of the full curvature tensor of a manifold, and can be compared to the metric tensor g. As such, bounding the Ricci tensor below by some scalar multiple of g (R ≥ kg, k ∈ R) is useful for uncovering analytic and geometric information about a manifold [1]. An example is that if R ≥ kg for k > 0, then the manifold M is compact. If R ≥ 0, the volume growth on (M, g)

n ωn−1 is at most Euclidean: ∀r > 0,V (x, r) ≤ Ωnr . Note that Ωn = n , where ωn−1 is the surface measure of the unit sphere in Rn. In addition, Ricci curvature can provide insight to what Sobolev inequalities a manifold satisﬁes. For example, [1, Theorem 3.3.8] shows that complete n-dimensional manifolds with nonnegative Ricci curvature and maximal volume growth satisfy the psuedo-Poincar´e inequality:

Theorem 4.1: Let (M, g) be a complete manifold of dimension n with Ricci curvature R ≥ 0. Assume that there exists c > 0 such that ∀r > 0,V (x, r) ≥ crn. Then,

∞ ∀r > 0, ∀f ∈ C0 (M), ||f − fr||1 ≤ (Ωn/c)r||∇f||1.

This is the psuedo-Poincar´einequality on M. Furthermore, all Sobolev inequalities θ (Sr,s(p)) with 1/r = θ(1/p = 1/n) + (1 + θ)/s, 0 < r, s ≤ ∞, θ ∈ (0, 1] are satisﬁed on M. Particularly, for all p ∈ [1, n),

∞ ∀f ∈ C0 (M), ||f||p∗ ≤ C(n, p)||∇f||p.

19 ∞ Proof. Observe that for f ∈ C0 (M), Z Z 1 ||f − fr||1 = f(x) − f(y)dy dx M V (x, r) B(x,r) Z Z 1 (y) ≤ |f(x) − f(y)| B(x,r) dydx. M M V (x, r)

Note that 1 is the indicator function; 1A(x) = 1 if x ∈ A, and 1A(x) = 0 if x 6∈ A. In the proof of this theorem, we find Z Z Z r Z ρ √ |f(x) − f(y)|dy ≤ |∇f(t, θ)|dt g(ρ, θ)dρdθ B(x,r) Sn−1 0 0 by integrating along the geodesic segment from x to y using polar exponential coordinates y = √ (ρ, θ) around x. The Riemannian volume element in polar coordinates is dy = g(ρ, θ)dρdθ. Now, we use the hypothesis that (M, g) has non-negative Ricci curvature: by Bishop’s √ theorem, the function s 7→ g(s, θ)/sn−1 is non-increasing. Thus, we find Z Z Z r Z ρ √ |f(x) − f(y)|dy ≤ |∇f(t, θ)|t1−n g(t, θ)dtρn−1dρdθ B(x,r) Sn−1 0 0 rn Z Z r √ ≤ |∇f(t, θ)|t1−n g(t, θ)dtdθ n Sn−1 0 rn Z ∇f(y) = n−1 dy n B(x,r) d(x, y) Now, using this along with the hypothesis of maximal volume growth that V (x, r) ≥ crn, we find Z Z Z Z 1B(x,r)(y) 1 |∇f(y)| |f(x) − f(y)| dydx ≤ n−1 dydx M M V (x, r) cn M B(x,r) d(x, y) 1 Z Z 1 ≤ |∇f(y)| n−1 dxdy cn M B(y,r) d(x, y) √ By Bishop’s theorem, g(t, θ) ≤ tn−1, so we see Z 1 n−1 dx ≤ ωn−1r B(y,r) d(x, y) and so ω ||f − f || ≤ n−1 r||∇f|| . r 1 cn 1

Recalling that Ωn = ωn−1/n, this is the desired inequality, and concludes the proof of Theo- rem 4.1.

In fact, we can construct a psuedo-Poincar´einequality on M without maximal volume growth, as proven in [1, Theorem 3.3.9].

20 4.1 Bounding the Number of Ends

Topologically, an “end” of a manifold is a connected component of the ideal boundary of the manifold. An end of a manifold is a topologically distinct way to move to infinity on the manifold. If M is a compact manifold with a boundary, the number of ends in the interior of M is the number of connected components of ∂M. In their paper Positive Solutions to SchrödingerEquations and Geometric Applications, authors Munteanu, Schulze and Wang demonstrate that positive solutions to given Shrödingerequations on a manifold “must be of polynomial growth of fixed order under a suitable scaling invariant Sobolev inequality” [4]. Each end of a complete Riemannian manifold has an associated positive solution to this given Schrödingerequation; as such, we can use Sobolev inequalities to prove that a manifold’s number of ends is finite, and even estimate that number. We will discuss two theorems from this paper that demonstrate how Sobolev inequalities can be used to bound a manifold’s number of ends.

Theorem 4.2 (cf. [4, Theorem 1.2]): Let (M, g) be a complete Riemannian manifold of dimension n ≥ 3 that satisﬁes the inequality

n−2 Z 2n n Z ∞ n−2 2 2 ∀f ∈ C0 (M), f ≤ A (|∇f| + σf ) M M for A > 0 a constant and σ ≥ 0 a continuous function. Set Z 1 2 n−1 α = lim sup (r σ) 2 < 0 R→∞ V (p, R) B(p,R) and V (p, R) V∞ = lim sup n < ∞, R→∞ R with p ∈ M a ﬁxed point, and r(x) = d(p, x) the distance function to p. Then the number of ends of M is bounded above by a constant Γ = Γ(n, A, α, V∞) dependent solely on n, A, α, and V∞. Recall from earlier that B(p, R) denotes the ball of radius R centered at p, and V (p, R) denotes the measure of this ball. Below is a more general version of Theorem 4.2.

Theorem 4.3 (cf. [4, Theorem 2.1]): Let (M, g) be an n-dimensional complete Riemannian manifold satisfying the Sobolev inequality

n−q Z qn n Z ∞ n−q q q ∀f ∈ C0 (M), |f| ≤ A (|∇f| + σ|f| ) (9) M M

21 for some q ∈ [1, n − 1], with A > 0 a constant and σ ≥ 0 a continuous function. Set Z 1 q n−1 α = lim sup (r σ) q R→∞ V (p, R) B(p,R)

and V (p, R) V∞ = lim sup n . R→∞ R

If both α and V∞ are ﬁnite, then the number of ends of M is bounded above by a constant

Γ(n, A, α, V∞) dependent solely on n, A, α, and V∞.

Proof. For an end E of an n-dimensional complete Riemannian manifold M, write E(R) = B(p, R) ∩ E. Let M have at least k > 1 ends, and take R such that

k [ B(p, 2R) \ B(p, R) = Ei(2R) \ Ei(R). i=1

V (p,t) By the deﬁnition of V∞, we know that tn ≤ 2V∞, and by the deﬁnition of α we see

k Z X q n−1 (r σ) q ≤ 2αV (p, 3R) i=1 Ei(3R)\Ei(R)

k Z X n−1 V (p, 3R) =⇒ σ q ≤ C 0 Rn−1 i=1 Ei(3R)\Ei(R)

The constant C0 is dependent on n, A, α,, and V∞, as with the C1,C2... seen below.

Label the ends E1, ..., Ek such that

Z n−1 σ q Ei(3R)\Ei(R) i=1,...,k is an increasing sequence. Then, we see Z n−1 2C0V (p, 3R) q σ ≤ n−1 , (10) Ei(3R)\Ei(R) kR

k k for all i ∈ 1, 2 . For all i ∈ 1, 2, ..., 2 pick a point

zi ∈ ∂Ei(2R) and relabel E1, ..., E k if necessary such that b 2 c V (zi,R) k i=1,...,b 2 c

22 is an increasing sequence. Assume for a contradiction that C V (z ,R) ≥ 1 Rn 1 k

n+2 where C1 = 3 V∞. We know that

B(zi,R) ⊂ Ei(3R) \ Ei(R)

k b 2 c and that B(zi,R) i=1 are disjoint in B(p, 3R). By our assumption, we see

b k c 2 X k C1 C1 V (p, 3R) ≥ V (z ,R) ≥ Rn ≥ Rn = 3V (3R)n. i 2 k 3 ∞ i=1

V (p,t) However, because tn ≤ 2V∞, we know that

n V (p, 3R) ≤ 3V∞(3R) ,

and so this is a contradiction. Thus, we see C V (z ,R) < 1 Rn. (11) 1 k

C1 n Let E = E1, z = z1, so z ∈ ∂E(2R) and V (z, R) < k R . By (10) we know Z n−1 C2V (p, 3R) q σ ≤ n−1 . E(3R)\E(R) kR

Let γ(t) be a minimizing geodesic from p to z, with γ(0) = p, γ(2R) = z: t ∈ [0, 2R]. For 2R t ∈ [4R/3, 5R/3] and x ∈ γ(t), because d(x, z≤ 3 the triangle inequality implies R B x, ⊂ B(z, R). 3

Thus, we see by (11) that

R C V x, < 1 Rn (12) 3 k

for all x = γ(t) for t ∈ [4R/3, 5R/3]. Next, authors Munteanu, Schulze, and Wang assume for a contradiction that

Z q n−q−1 σ ≤ δr n−1 V (x, r) n−1 (13) B(x,r)

23 for r ∈ (0,R/3) with δ > 0 a constant. Fix an r ∈ (0,R/3) and consider the cut-oﬀ function 2 φ with support in B(x, r) such that φ = 1 on B(x, r/2) and |∇φ| ≤ r . Applying the Sobolev inequality (9), we see

n−q r n 2q Z V x, ≤ A q V (x, r) + σ . 2 r B(x,r) By the assumption (13), we see

n−q q r n 2 q n−q−1 V x, ≤ A V (x, r) + δr n−1 V (x, r) n−1 2 rq

for any r ∈ (0,R/3). Assume that there exists an r0 ∈ (0,R/3) such that V (x, r0) ≤ n−1 q n−q δ r0 . This brings us to

n−q n r0 q n−1 n−q V x, ≤ A(2 + 1)δ q r 2 0 n r0 n q n n−1 + n−1 r0 =⇒ V x, ≤ 2 (A(2 + 1)) n−q δ n−q q 2 2 Choose δ such that n q n n−1 2 (A(2 + q)) n−q δ n−q < 1.

Then, we have n r0 n−1 r0 V x, ≤ δ q . 2 2

Thus, given the assumption that (13) holds for r ∈ (0,R/3), then if there exists an r0 in this n−1 q n−q interval such that V (x, r0) ≤ δ r0 , the above inequality holds. If k is large enough to n n−1 3 C1 q satisfy k ≤ δ , (12) implies that n R n−1 R V x, < δ q 3 3 and we can set r0 = R/3 to ﬁnd

n R n−1 R V x, ≤ δ q 3 · 2m 3 · 2m

n−1 q by induction. However, if we set δ small enough that δ < V (0, 1)Rn , then as m → ∞ this is a contradiction. Therefore, (13) does not hold; for any x = γ(t) with t ∈ [4R/3, 5R/3] there exists an rx ∈ (0,R/3) such that

Z q n−q−1 n−1 σ > δrx V (x, rx) n−1 B(x,rx)

24 and the H¨olderinequality brings us to

q Z Z n−1 n−1 n−q−1 σ ≤ σ q V (x, rx) n−1 B(x,rx) B(x,rx)

Z n−1 1 =⇒ σ q ≥ rx B(x,rx) C3 for any x = γ(t) for t ∈ [4R/3, 5R/3]. A covering argument as in [4] implies that we can choose at most countably many disjoint balls {B(xm, rxm )}m≥1 with xm = γ(tm), tm ∈ [4R/3, 5R/3], each satisfying the above inequality. These balls cover at least one third of the geodesic γ([4R/3, 5R/3]). Thus, X 15R 4R R r ≥ − = xm 3 3 3 9 m≥1 and so Z Z R X X n−1 n−1 ≤ r ≤ C σ q ≤ C σ q 9 xm 3 3 m≥1 m≥1 B(xm,rxm B(z,R) R because B(xm, rxm ) m≥1 are disjoint in B(z, R) and B x, 3 ⊂ B(z, R). The authors conclude that Z 1 n−1 C2V (p, 3R) q R ≤ σ ≤ n−1 9C3 E(3R)\E(R) kR kRn k =⇒ V (p, 3R) ≥ = Rn. 9C3C2 C4 n If k > 2V∞C43 , we have n V (p, 3R) > 2V∞(3R)

V (p,t) This is a contradiction of tn ≤ 2V∞ (from the deﬁnition of V∞). Thus, k, the number of n ends of M, is bounded above above by 2V∞C43 . This proves the theorem.

In addition, there is a corollary to this theorem that uses the mean curvature H of the manifold M.

Corollary 4.4: If M n is a complete submanifold of RN with n ≥ 2, let 1 Z α˜ = lim sup (r|H|)n−1 < ∞ R→∞ V (p, R) B(p,R) and V (p, R) V∞ = lim sup n < ∞. R→∞ R Then the number of ends of M is bounded above by a constant Γ dependent only on n, α,˜

and V∞.

25 5 Applications of Sobolev Inequalities on Manifolds

When physicists and mathematicians study the conduction and diffusion of heat, a heat kernel is the fundamental solution to the heat equation when operating in a specified domain with boundary conditions. The heat kernel represents the change or evolution in temperature in a region when this region’s boundary is held at a fixed temperature. Sobolev inequalities, notably the Nash inequality, have been used to study the heat dif- −t∆ fusion semigroup Ht = e . In fact, the Nash inequality implies that Ht is a bounded operator from L1 to L∞ with explicit time dependent estimates. The following theorem [1, Theorem 4.1.1] was first proven by Nash and contributes to this connection between Sobolev inequalities and the heat kernel.

Deﬁnition 5.1: The q to p norm, notated ||A||q→p, is deﬁned as

||A||q→p := max{||Ax||p : ||x||q = 1}.

Here, the x represent elements of the set that the operator A acts on in-context. In our context, we will be considering the heat diﬀusion semigroup Ht as our operator, with our elements being functions f ∈ D, the domain of a Dirichlet form on L2(M, µ).

Theorem 5.2: Let Q be a Dirichlet form on L2(M, µ) with domain D and associated semigroup (Ht)t>0. Assume that the Nash inequality

2(1+2/v) 4/v ∀f ∈ D, ||f||2 ≤ CQ(f, f)||f||1 is satisﬁed for some v > 0. Then (Ht)t>0 is ultracontractive and

v/2 ∀t > 0, ||Ht||1→∞ ≤ (Cv/2t) , or, equivalently, (Ht)t>0 admits a density with respect to µ satisfying

∀t > 0, sup {h(t, x, y)} ≤ (Cv/2t)v/2, x,y∈M where h(t, x, y) is the non-negative smooth function known as the heat kernel associated to V −1∆ on L2(M, dµ). Equipped with this theorem, we move forward to setting Gaussian estimates on the heat kernel. Let M be a complete non-compact Riemannian manifold with Riemannian measure

26 µ. Let ∆ be the Laplacian on M and let h(t, x, y) be the heat diﬀusion kernel on M; for each x ∈ M, h(t, x, y) = u(t, y) is the minimal solution of  (∂t + ∆)u = 0

u(0, y) = ∂x(y).

This means that h(t, x, y) is the kernel of the semigroup Ht associated to the Dirichlet R 2 1 ∞ form Q(f, f) = |∇f| dµ with domain W2 (M): the closure of C0 (M) under the norm p 2 2 ||f||2 + ||∇f||2. ∞ For any function φ ∈ C0 (M) with ||∇φ||∞ ≤ 1 and any complex number α, consider the semigroup Z α,φ −αφ(x) αφ(y) −αφ(x) αφ Ht f(x) = e h(t, x, y)e f(y)dy = e Ht(e f)(x);

this is a semigroup of operators on the spaces Lp(M, µ), and its generator is given by

−αφ α,φ −Aα,φf = −e ∆(e f).

α,φ The semigroup (Ht )t>0 satisﬁes

α,φ α2t ∀t > 0, ||Ht ||2→2 ≤ e .

In addition, if α is real, then for all t > 0, ζ = eir with |r| ≤ , ∈ (0, π/4] we have

α,φ α2(1+)t ||Htζ ||2→2 ≤ e

and there exists a constant C such that eα2(1+)t ||∂kHα,φ|| ≤ Ckk! . t t 2→2 (t)k

v/4 If the heat diﬀusion semigroup (Ht)t>0 satisﬁes ||Ht||2→∞ ≤ (C/t) for all t > 0 for some constant C, then for all p ∈ [2, ∞]

v/2p ∀t > 0, ||Ht||p→∞ ≤ (C/t) .

In addition, if there exists a constant C and v > 0 such that

∀t > 0, ||Ht||2→∞≤(C/t)v/4 ,

then α,φ v(1/p−1/q)/2 qα2t/2 ||Ht ||p→q ≤ (C/t) e

27 ∞ for all q ≥ p ≥ 2, t > 0, α ∈ R and φ ∈ C0 (M) with ||∇φ||∞ ≤ 1. Now, we will consider [1, Theorem 4.2.6].

Theorem 5.3: Assume that a manifold M satisﬁes the Nash inequality

∞ 2(1+2/v) 2 4/v ∀f ∈ C0 (M), ||f||2 ≤ C||∇f||2||f||1 . (14)

Then, for any δ > 0 there exists a finite constant C(δ) such that the kernel h(t, x, y) of the −t∆ heat diffusion semigroup Ht = e , t > 0 satisfies d(x, y)2 h(t, x, y) ≤ (C(δ)/t)v/2 exp . 4(1 + δ)t

This theorem states that if a manifold M satisfies the Nash inequality, we have an upper bound on the heat kernel of the heat diffusion semigroup Ht. However, this Gaussian upper bound does not make for the best estimate due to the exponential growth. We want to find a way to refine this estimate. If M is a Riemannian manifold of dimension n that satisfies the Nash inequality (14) for some v > 0, we have seen that the volume lower bound ∀t > 0,V (t) ≥ ctv

is implied, and so v ≥ n. If the volume growth is faster than tv, this bound is not useful. So, we have the following theorem [1, Theorem 4.2.8], that has an added restriction on volume growth to provide us with bounds above and below the heat kernel.

Theorem 5.4: Fix v > 0. Assume that M satisﬁes the Nash inequality (14) and the volume growth condition −v ∀x ∈ M, ∀r > 0, c0 ≤ r V (x, r) ≤ C0.

Then the heat kernel h(t, x, y) is bounded above and below on the diagonal by

ct−v/2 ≤ h(t, x, x) ≤ Ct−v/2.

The Nash inequality, a Sobolev-type inequality, is integral to these theorems that bound the heat kernel.

28 References

[1] Laurent Saloﬀ-Coste. Aspects of Sobolev-Type Inequalities. London Mathematical Soci- ety Lecture Note Series No. 239, Cambrdige University Press, 2002.

[2] Thayer Watkins. Area-Volume Formulas for N-Dimensional Spheres and Balls. https://www.sjsu.edu/faculty/watkins/ndim.htm

[3] John Volker. Numerical Mathematics 3: Chapter 3 - Introduction to Sobolev Spaces. Weierstrass Institute for Applied Analysis and Stochastics. 2018. https://www.wias-berlin.de/people/john/LEHRE/lehrealt.html

[4] Ovidiu Munteanu, Felix Schulze, Jiaping Wang. Positive Solutions to Schr¨odingerEqua- tions and Geometric Applications. Crelle’s Journal. 2020.