4 Metric Spaces

Let X be a nonempty set. A distance function or a metric on a set X is a function which assigns to a pair of points x and y in X a real number d(x,y) having the following properties:

M1 d(x,y) ≥ 0 and d(x,y) = 0 if and only if x = y.

M2 d(x,y)= d(y,x) for all x,y ∈ X.

M3 (triangle inequality): d(x, z) ≤ d(x,y)+ d(y, z) for all x,y and z ∈ X.

Deﬁnition 4.1. A metric space is a pair (X, d) where d is a metric deﬁned on the set X.

Given a ∈ X and r> 0, the sets

B(a, r)= BX (a, r)= {x ∈ X| d(a, x) < r}

B(a, r)= BX (a, r)= {x ∈ X| d(a, x) ≤ r} are called the open ball and the closed ball with center at a and radius r.

Example 4.2. The set of real numbers R is a metric space with the standard metric d(x,y)= |x − y|, x,y ∈ R. The open ball Br(a) with center at a and radius r> 0 is equal to the open interval (a − r, a + r).

Example 4.3. Let X be any nonempty set and let d : X × X → R be deﬁned by 0 x = y, d(x,y)= (1 x 6= y. Then d is a metric called the discrete metric on X.

Example 4.4. Let (X, d) be a metric space and Y is a nonempty subset of X. Then the restriction of d to Y × Y , dY = d|Y × Y , is a metric on Y . It is called the induced metric. The pair (Y, dY ) is a metric space, metric subspace of X. (We will refer to Y as a subspace of X, rather than (Y, dY ) as a subspace of (X, d).) The open ball and closed balls in (Y, dY ) are

BY (a, r)= BX (a, r) ∩ Y and BY (a, r)= BX (a, r) ∩ Y.

15 Example 4.5. Let (Xi, di), 1 ≤ j ≤ n, be metric spaces and let X = X1 ×. . .×Xn. Then each of the following functions deﬁnes a metric (product metric) on X: n d(x,y)= di(xi,yi) i=1 X 1 2 n / 2 d(x,y)= di(xi,yi) " =1 # Xi d(x,y)= max dij(xi,yj) 1≤j≤n where x = (x1,...,xn), y = (y1,...,yn) in X. The pair (X, d) deﬁned above is called the metric product or just a product) of (Xi, di). The open ball and closed balls in X are

BX (a, r)= BX1 (a1, r) × . . . × BXn (an, r)

BX (a, r)= BX1 (a1, r) × . . . × BXn (an, r) where a = (a1,...,an) ∈ X. A subset U of a metric space X is called a neighborhood of a ∈ X, if there is r> 0 such that B(a, r) ⊆ U.

Example 4.6. Clearly, B(a, r) and B(a, r) are neighborgoods of a. If U1, U2 are neighborhoods of a, then their union and intersection are neighborhoods of a. Indeed, let B(a, rj) ⊆ Uj for j = 1, 2. Then B(a, r) ⊆ U1 ∩U2 ⊆ U1 ∪U2 where r = min{r1, r2}.

4.1 Norms and normed vector spaces We next deﬁne the class of metric spaces which are the most interesting in analysis. Let X be a vector space over R. A norm is a function k·k : X → [0, ∞) having the following properties.

N1 kxk = 0 if and only if x = 0.

N2 kαxk = |α| · kxk for all x ∈ X and α ∈ R.

N3 kx + yk ≤ kxk + kyk for all x,y ∈ X.

A pair (X, k·k) is called a normed vector space.

16 Proposition 4.7. Let (X, k·k) be a normed space. Then the function

d : X × X → R, d(x,y)= kx − yk is a metric on X. It is called the metric induced from the norm. Proof. The axioms M1 and M2 are clear. If x,y and z ∈ X, then, in view of N3,

d(x, z)= kx − zk = k(x − y) + (y − z)k ≤ kx − yk + ky − zk = d(x,y)+ d(y, z), and so the triangle inequality follows.

Example 4.8. The absolute value |·| on R is a norm.

Example 4.9 (Induced norm). Let Y be a nonempty subspace of a normed space (X, k·k). Then the restriction k·kY = k·k |Y is a norm on Y . It is called the induced norm on Y .

Example 4.10 (Product norm). Let (X1, k·k1),..., (Xn, k·kn) be normed spaces over R and let X = X1 ×. . .×Xn. Then each of the following function deﬁnes a norm on X: n

kxk := kxiki i=1 X 1 2 n / 2 kxk := kxiki " =1 # Xi kxk = max{kx1k1 ,..., kxnkn} where x = (x1,...,xn) ∈ X. Either of them is called the product norm on X.

Example 4.11 (Space of Bounded Functions). Let X be a nonempty set and let (Y, k·k) be a normed space over R. A function f : X → Y is called bounded if its image is bounded in Y , i.e., f(X) ⊂ BY (0, r) for some r > 0. Explicitly, this means that there is r > 0 so that kf(x)k < r for all x ∈ X. The set of all bounded function f : X → Y is denoted by B(X,Y ). This is a vector space, called space of bounded functions from X to Y . If f ∈ B(X,y), deﬁne

kfk∞ = sup{kf(x)k | x ∈ X}. (1)

17 If f, g ∈ B(X,Y ) and x ∈ X, then

kf(x)+ g(x)k ≤ kf(x)k + kg(x)k ≤ kfk∞ + kgk∞ and since x was arbitrary element in X one concludes that

kf + gk∞ ≤ kfk∞ + kgk∞ .

Hence k·k∞ satisﬁes the triangle inequality. The other axioms are also sat- isﬁed. So, (B(X,Y ), k·k∞) is a normed space.

4.2 Inner product spaces Let X be a vector space over R. Then a function h·, ·i : X ×X → K is called the inner product or scalar product, if it satisﬁes the following conditions.

IP1 hx,xi≥ 0 and hx,xi = 0 if and only if x = 0.

IP2 hαx + βy,zi = αhx, zi + βhy, zi, for all x,y,z ∈ X and all α, β ∈ R.

IP3 hx,yi = hy,xi for all x,y ∈ X.

A pair (X, h·, ·i) is called an inner product space.

n Example 4.12. Consider X = R and for x = (x1,...,xn),y = (y1,...,yn) in Rn let n hx,yi = xiyi. =1 Xi Then h·, ·i is an inner product on Rn. It is called the standard inner product.

Proposition 4.13 (Cauchy-Schwarz Inequality). Let (X, h·, ·i) be an inner product space. Then

1 2 1 2 |hx,yi| ≤ hx,xi / hy,yi / for all x,y ∈ X with the equality if and only if x and y are linearly dependent.

Proof. Fix two points x,y ∈ X. Without loss of generality we may assume that y 6= 0 (if y = 0 then the claim follows since both sides are equal to 0). For every t ∈ R, we have

2 0 ≤ hx − ty,x − tyi = hx,xi− 2thx,yi + t hy,yi.

18 The right-hand side deﬁnes a quadratic function f(t) = hx,xi− 2thx,yi + 2 hx,yi 0 0 t hy,yi. The function f has a minimum at t = hy,yi . Evaluating f(t ) we get

2 hx,yi hx,yi hx,yi2 0 ≤ f(t0)= hx,xi− 2 hx,yi + hy,yi = hx,xi− hy,yi hy,yi hy,yi implying that hx,yi2 ≤ hx,xihy,yi, that is |hx,yi| ≤ hx,xi1/2hy,yi1/2.

Corollary 4.14 (Classical Cauchy-Schwarz inequality). If x = (x1,...,xn), n y = (y1,...,yn) ∈ R , then

2 n n n 2 2 x y ≤ |x | |y | . j j j j=1 j=1 j=1 X X X

Proposition 4.15. Let (X, h· , ·i) be an inner product space and let kxk = hx,xi, x ∈ X.

Then k·k is a norm on X, the normp induced from the inner product.

Proof. It suﬃces to prove the triangle inequality. Let x,y and z ∈ X. For x,y ∈ X, we have

2 kx + yk = hx + y,x + yi = hx,xi + 2hx,yi + hy,yi 2 2 ≤ kxk + 2 |hx,yi| + kyk 2 2 2 ≤ kxk + 2 kxk kyk + kyk = (kxk + kyk)

So, kx + yk ≤ kxk + kyk as claimed.