Solutions of Some Exercises In this section the formulas are numbered (S1), (S2), etc, in order to avoid any confusion with formulas from the previous sections. 1.1 1. The equality f, x = x 2 implies that x ≤ f . Corollary 1.3 implies that F(x)is nonempty. It is clear from the second form of F(x)that F(x)is closed and convex. 2. In a strictly convex normed space any nonempty convex set that is contained in a sphere is reduced to a single point. 3. Note that 1 1 f, y ≤ f y ≤ f 2 + y 2. 2 2 Conversely, assume that f satisfies 1 1 (S1) y 2 − x 2 ≥ f, y − x ∀y ∈ E. 2 2 First choose y = λx with λ ∈ R in (S1); by varying λ one sees that f, x = x 2. Next choose y in (S1) such that y =δ>0; it follows that 1 1 f, y ≤ δ2 + x 2. 2 2 Therefore we obtain 1 1 δ f = sup f, y ≤ δ2 + x 2. y∈E 2 2 y =δ The conclusion follows by choosing δ = x . 4. If f ∈ F(x)one has H. Brezis, Functional Analysis, Sobolev Spaces and Partial Differential Equations, 371 DOI 10.1007/978-0-387-70914-7, © Springer Science+Business Media, LLC 2011 372 Solutions of Some Exercises 1 1 y 2 − x 2 ≥ f, y − x 2 2 and if g ∈ F(y)one has 1 1 x 2 − y 2 ≥ g,x − y . 2 2 Adding these inequalities leads to f − g,x − y ≥0. On the other hand, note that f − g,x − y = x 2 + y 2 − f, y − g,x ≥ x 2 + y 2 − 2 x y . 5. By question 4 we already know that x = y . On the other hand, we have F(x)− F(y),x − y =[ x 2 − F(x),y ]+[ y 2 − F(y),x ], and both terms in brackets are ≥ 0. It follows that x 2 = y 2 = F(x),y = F(y),x , which implies that F(x) ∈ F(y)and thus F(x) = F(y)by question 2. 1.2 1(a). f E = max |fi|. 1≤i≤n 1(b). f ∈ F(x)iff for every 1 ≤ i ≤ n one has (sign xi) x 1 if xi = 0, fi = anything in the interval [− x 1, + x 1] if xi = 0. 2(a). n f E = |fi|. i=1 2(b). Given x ∈ E consider the set I ={1 ≤ i ≤ n;|xi|= x ∞}. Then f ∈ F(x)iff one has (i) fi = 0 ∀i/∈ I, ≥ ∀ ∈ | |= (ii) fixi 0 i I and i∈I fi x ∞. 3. n 1/2 2 f E = |fi| i=1 and f ∈ F(x)iff one has fi = xi ∀i = 1, 2,...,n. More generally, Solutions of Some Exercises 373 n 1/p p f E = |fi| , i=1 p−2 p−2 where 1/p + 1/p = 1, and f ∈ F(x) iff one has fi =|xi| xi/ x p ∀i = 1, 2,...,n. 1.3 = α = + ∀ 1. f E 1 (note that f(t ) 1/(1 α) α>0). 1 − = ≡ 2. If there exists such a u we would have 0 (1 u)dt 0 and thus u 1; absurd. 1.5 1. Let P denote the family of all linearly independent subsets of E. It is easy to see that P (ordered by the usual inclusion) is inductive. Zorn’s lemma implies that P has a maximal element, denoted by (ei)i∈I , which is clearly an algebraic basis. Since ei = 0 ∀i ∈ I, one may assume, by normalization, that ei =1 ∀i ∈ I. 2. Since E is infinite-dimensional one may assume that N ⊂ I. There exists a (unique) linear functional on E such that f(ei) = i if i ∈ N and f(ei) = 0if i ∈ I\N. 3. Assume that I is countable, i.e., I = N. Consider the vector space Fn spanned by ∞ = (ei)i≤i≤n. Fn is closed (see Section 11.1) and, moreover, n=1 Fn E. It follows =∅ from the Baire category theorem that there exists some n0 such that Int(Fn0 ) . = Thus E Fn0 ; absurd. 1.7 1. Let x,y ∈ C, so that x = lim xn and y = lim yn with xn,yn ∈ C. Thus tx + (1 − t)y = lim[txn + (1 − t)yn] and therefore tx + (1 − t)y ∈ C ∀t ∈[0, 1]. Assume x,y ∈ Int C, so that there exists some r>0 such that B(x,r) ⊂ C and B(y,r) ⊂ C. It follows that tB(x, r) + (1 − t)B(y,r) ⊂ C ∀t ∈[0, 1]. But tB(x, r) + (1 − t)B(y,r) = B(tx + (1 − t)y,r) (why?). 2. Let r>0 be such that B(y,r) ⊂ C. One has tx + (1 − t)B(y,r) ⊂ C ∀t ∈[0, 1], and therefore B(tx+(1−t)y,(1−t)r) ⊂ C. It follows that tx+(1−t)y ∈ Int C ∀t ∈[0, 1). ∈ ∈ = [ − 1 + 1 ] 3. Fix any y0 Int C.Givenx C one has x limn→∞ (1 n )x n y0 . But − 1 + 1 ∈ ∈ ⊂ (1 n )x n y0 Int C and therefore x Int C. This proves that C Int C and hence C ⊂ Int C. 1.8 1. We already know that 374 Solutions of Some Exercises p(λx) = λp(x) ∀λ>0, ∀x ∈ E and p(x + y) ≤ p(x) + p(y) ∀x,y ∈ E. It remains to check that (i) p(−x) = p(x) ∀x ∈ E, which follows from the symmetry of C. (ii) p(x) = 0 ⇒ x = 0, which follows from the fact that C is bounded. More precisely, let L>0 be such that x ≤L ∀x ∈ C. It is easy to see that 1 p(x) ≥ x ∀x ∈ E. L √ = + 2. C is not bounded. Consider for example the sequence/ un(t) n/(0 1 nt) and √ 1/2 ∈ = = 1 | |2 check that un C, while un n. Here p(u) 0 u(t) dt is a norm that is not equivalent to u . 1.9 1. Let n n P = λ = (λ1,λ2,...,λn) ∈ R ; λi ≥ 0 ∀i and λi = 1 , i=1 n so that P is a compact subset of R and Cn is the image of P under the continuous → n map λ i=1 λixi. 2. Apply Hahn–Banach, second geometric form, to Cn and {0}. Normalize the linear functional associated to the hyperplane that separates Cn and {0}. 4. Apply the above construction to C = A − B. 1.10 (A) ⇒ (B) is obvious. (B) ⇒ (A). LetG be the vector space spanned by the xi’s (i ∈ I).Givenx ∈ G = = write x i∈J βixi and set g(x) i∈J βiαi. Assumption (B) implies that this definition makes sense and that |g(x)|≤M x ∀x ∈ G. Next, extend g to all of E using Corollary 1.2. 1.11 (A) ⇒ (B) is again obvious. (B) ⇒ (A). Assume first that the fi’s are linearly independent (1 ≤ i ≤ n). Set n n α = (α1,α2,...,αn) ∈ R . Consider the map ϕ : E → R defined by ϕ(x) = ( f1,x ,..., fn,x ) . Let C ={x ∈ E; x ≤M + ε}. One has to show that α ∈ ϕ(C). Suppose, by contradiction, that α/∈ ϕ(C) and separate ϕ(C) and {α} (see Exercise 1.9). n Hence, there exists some β = (β1,β2,...,βn) ∈ R ,β = 0, such that ) * β · ϕ(x) ≤ β · α ∀x ∈ C, i.e., βifi,x ≤ βiαi ∀x ∈ C. Solutions of Some Exercises 375 It follows that (M +ε) βifi ≤ βiαi. Using asumption (B) one finds that βifi = 0. Since the fi’s are linearly independent one concludes that β = 0; absurd. In the general case, apply the above result to a maximal linearly independent subset of (fi)1≤i≤n. 1.15 1. It is clear that C ⊂ C and that C is closed. Conversely, assume that x0 ∈ C and x0 ∈/ C. One may strictly separate {x0} and C, so that there exist some f0 ∈ E and some α0 ∈ R such that f0,x <α0 < f0,x0 ∀x ∈ C. Since 0 ∈ C it follows that α0 > 0; letting f = (1/α0)f0, one has f, x < 1 < f, x0 ∀x ∈ C. Thus f ∈ C and we are led to a contradiction, since x0 ∈ C . 2. If C is a linear subspace then ⊥ C ={f ∈ E; f, x =0 ∀x ∈ C}=C . 1.18 −b if f = a, (a) ϕ(f ) = +∞ if f = a. ⎧ ⎨⎪f log f − f if f>0, (b) ϕ(f ) = 0iff = 0, ⎩⎪ +∞ if f<0. (c) ϕ(f ) =|f |. (d) ϕ(f ) = 0. +∞ if f ≥ 0, (e) ϕ(f ) = −1 − log |f | if f<0. (f) ϕ(f ) = (1 + f 2)1/2. 1 f 2 if |f |≤1, (g) ϕ(f ) = 2 +∞ if |f | > 1. = 1 | |p 1 + 1 = (h) ϕ (f ) f with 1. p p p 0if0≤ f ≤ 1, (i) ϕ(f ) = +∞ otherwise. 376 Solutions of Some Exercises 1 p f if f ≥ 0, (j) ϕ(f ) = p 0iff<0. +∞ ≥ if f 0, (k) ϕ (f ) = − 1 | |p p f if f<0. 1 (l) ϕ(f ) =|f |+ |f |p . p p 1 + 1 = 1.20 The conjugate functions are defined on with p p 1by ∞ ∞ 1 1 |f |2 if 1 |f |2 < +∞, (a) ϕ(f ) = 4 k=1 k k k=1 k k +∞ otherwise. +∞ +∞ a |f |k/(k−1) if a |f |k/(k−1) < +∞, (b) ϕ(f ) = k=2 k k k=2 k k +∞ otherwise, (k − ) = 1 with ak − .