Functional Analysis

Minor Test I: Hints to Solutions

1. For a ∈ U (0, 1), and x, y ∈ X

( )
dist x, U (0, 1) ≤ ∥x − a∥ ≤ ∥x − y∥ + ∥y − a∥.

Since the above inequality is true for any a ∈ U (0, 1),

( ) ( )
dist x, U (0, 1) ≤ ∥x − y∥ + dist y, U (0, 1) .

That is


( ) ( )
dist x, U (0, 1) − dist y, U (0, 1) ≤ ∥x − y∥. (∗)1

Similarly,
( ) ( )
dist y, U (0, 1) − dist x, U (0, 1) ≤ ∥x − y∥. (∗)2

From (∗)1 and (∗)2 , we have

( ) ( )
|f (x) − f (y)| = dist x, U (0, 1) − dist y, U (0, 1) ≤ ∥x − y∥.

Therefore, f : X → R is a Lipschitz map, and hence, in particular, uniformly continuous.

We know that for x ∈ X and A ⊆ X,

dist(x, A) = 0 ⇔ x ∈ Ā.

Therefore

{ ( ) }
ker(f ) := {x ∈ R2 : f (x) = 0} = Ū (0, 1) = {x ∈ R2 : ∥x∥ ≤ 1} = x = x(1), x(2) : |x(1)|+|x(2)| ≤ 1 .

tn ∥.∥∞
2. (a) False. Consider the sequence (xn ) in A, defined by xn (t) = n ,0 ≤ t ≤ 1. Then xn −−−→ 0, but

x ≡ 0 is not injective.
( ) ( )
(b) True. Since C[0, 1], ∥.∥∞ is a Banach space, it is enough to prove that B is closed in C[0, 1], ∥.∥∞ .
∥.∥∞
Let (xn ) be a sequence in B and xn −−−→ x. Note that convergence in the supnorm is equivalent to the

uniform convergence and uniform convergence implies point wise convergence. Therefore, the sequence
( )
xn ( 21 ) converges to x( 12 ). Since xn ( 12 ) = 0 for all n ∈ N, it follows that x( 12 ) = 0 and hence that B
( )
is closed in C[0, 1], ∥.∥∞ .

3. We have the following theorem regarding best approximation:


Theorem 1 Let X be a normed linear space and Y be a finite dimensional subspace of X. Then for

each x ∈ X, there exists x∗ ∈ Y such that

∥x − x∗ ∥ ≤ ∥x − u∥ ∀ u ∈ Y.

Further, if Y is a proper subspace, there exists x̃ ∈ X such that ∥x̃∥ = 1 and dist(x̃, Y ) = 1.