# Functional Analysis problems need checking

Question 1

Prove that if $(V, \|\cdot\|)$ is a normed vector space, then

$$\left| \|x\| - \|y\| \right| \leq \|x-y\|$$

for every $x,y \in V$. Then deduce that the norm is a continuous function from $V$ to $\mathbb{R}$.

Solution

Let $x = x-y+y$. Taking norms on both sides gives

$$\|x\| = \|x-y+y\| \leq \|x-y\| + \|y\|$$

This implies that

$$\|x\| - \|y\| \leq \|x-y\|$$ (1)

Now let $y = y -x + x$. Taking norms on both sides gives

$$\|y\| = \|y - x + x\| \leq \|y - x\| + \|x\|$$

Which implies that

$$\|y\| - \|x\| \leq \|y-x\|$$

Which is also

$$-\left(\|x\|-\|y\|\right) \leq \|x-y\|$$ (2)

By considering both (1) and (2) we have

$$\left| \|x\| - \|y\| \right| \leq \|x - y\|$$

For continuity, suppose we fix $y = x_0 \in V$. Then for every $x \in V$ the triangle inequality gives us

$$\left| \|x\| - \|x_0\| \right| \leq \|x - x_0\|$$

Which implies

$$\left| \|x\| - \|x_0\| \right| < \epsilon$$

Satisfying

$$\|x-x_0\| < \epsilon$$

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Galileo
Homework Helper
Looks good. If it's homework I suggest to write down the continuity proof more clearly and completely like you did with the first question.

Thankyou for replying Galileo. Here is my next question/solution...

Question 2

Let $\{a_n\}$ be a fixed bounded sequence of complex numbers, and define $T:l^2 \rightarrow l^2$ by

$$T(\{x_n\}) = \{a_nx_n\}$$

Prove that $T$ is linear and bounded with

$$\|T\| = \sup |a_n|$$

The first thing I did was assume that the norm is the usual one

$$\|x\| = \left(\sum_{n=1}^{\infty} |x_n|^2 \right)^{1/2}$$

Then to show that $T$ is linear I showed two things:

(1) $$T(\{x_n+y_n\}) = \{a_n(x_n+y_n)\} = \{a_nx_n + a_ny_n\} = \{a_nx_n\} + \{a_ny_n\} = T(\{x_n\}) + T(\{y_n\})$$

(2) $$T(\lambda\{x_n\}) = \{\lambda a_nx_n\} = \lambda \{a_n x_n\} = \lambda T(\{x_n\})$$

For every $x_n,y_n \in l^{\infty}$

Hence linear.

Further, for every $x_n \in l^{\infty}$ we have

$$\|T(\{x_n\})\|^2 = \sum_{i=1}^{\infty} |a_nx_n|^2 \leq \left(\sup_{n\in\mathbb{N}}|a_n|\right)^2\sum_{n=1}^{\infty}|x_n|^2 = \left(\sup_{n\in\mathbb{N}}|a_n|\right)^2\|x\|^2$$

Therefore $T:l^2\rightarrow l^2$ is bounded, with

$$\|T(\{x_n\})\| \leq M \|x\|$$

For all $\{x_n\} \in l^{\infty}$ where

$$M = \sup_{n\in\mathbb{N}}|a_n| = \|a_n\|_{\infty}$$

ie, the supremum norm of $\{a_n\} \in l^{\infty}$.

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MalleusScientiarum
Your first proof works for me just fine, but you have to use the given norm that tells you that
$$\| T(\{x_n\}) \| = \textrm{sup}(a_n)$$
This is a function mapping sequences to sequences, and you've been given a norm differing from the Euclidean norm, so you have to use that one.

If the norm I am actually meant to be using is

$$\|T\| = \sup |a_n|$$

Then isn't $T$ obviously bounded by the supremum of the $a_n's$? This is probably wrong, since Im used to seeing an inequality instead of an equality - which is alarming.

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MalleusScientiarum
Yeah, the norm should be bounded in the case of bounded coefficients.

How would I formalize this statement? Or is it enough to simply say that the norm is bounded by the supremum of the coefficients?

Im having some difficulties proving some basic properties of the adjoint operator. I want to prove the following things:

1) There exists a unique map $T^*:K\rightarrow H$
2) That $T^*$ is bounded and linear.
3) That $T \rightarrow K$ is isometric if and only if $T^*T = I$.
4) Deduce that if $T$ is an isometry, then $T$ has closed range.
5) If $S \in B(K,H)$, then $(TS)^* = S^*T^*$, and that $T^*^* = T$.
6) Deduce that if $T$ is an isometry, then $TT^*$ is the projection onto the range of $T$.

Note that $H,K$ are Hilbert Spaces.

There are quite a few questions, and I am hoping that by proving each one I will get a much better understanding of these adjoint operators. Now I think I have made a fairly good start with these proofs, so I'd like someone to check them please.

We'll begin with the first one.

1) (To prove that $T^*$ is unique I'll be referring to Riesz's Theorem.)

I want to prove that there exists a unique mapping $T^*:K\rightarrow H$ such that

$$\langle Th,k \rangle = \langle h, T^*k \rangle \quad \forall \, h \in H, \, k\in K$$

For each $k \in K$, the mapping $h \rightarrow \langle Th, k\rangle_K$ is in $H^*$. Hence by Riesz's theorem, there exists a unique $z \in H$ such that

$$\langle Th,k \rangle_K = \langle h,z \rangle_H \quad \forall \, h \in H$$.

Therefore there exists a unique map $T^*: K \rightarrow H$ such that

$$\langle Th, k\rangle_K = \langle h,T^*k\rangle_H \quad \forall \, h \in H, \, k \in K$$.

Therefore there exists a unique $T^*$. $\square$

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2a) To see that $T^*$ is linear, take $k_1, k_2 \in K$ and $\lambda \in \mathbb{F}$, then for any $h \in H$ we have

$$\langle Th,k_1+\lambda k_2 \rangle_K &=& \langle Th,k_1\rangle_K + \overline{\lambda}\langle Th, k_2 \rangle_K \\ &=& \langle h,T^*k_1\rangle_K + \overline{\lambda}\langle h,T^*k_2\rangle_K \\ &=& \langle h, T^*k_1 + \lambda T^*k_2\rangle_K$$

Hence

$$T^*(k_1+\lambda k_2) = T^*(k_1) + \lambda T^*(k_2)$$

$T^*$ is linear. $\square$

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2b) To prove that $T^*$ is bounded note first that

$$\|T^*k\|^2 = \langle T^* k, T^*k \rangle_K = \langle T(T^*k), k \rangle \leq \|T(T^*k)\|\|k\| \leq \|T\|\|T^*k\|\|k\| \quad \forall \, k \in K$$

Now suppose that $\|T^*k\| > 0$. Then dividing the above by $\|T^*k\|$ we have

$$\|T^*k\| \leq \|T\|\|k\| \quad \forall \, k \in K$$

Note that this is trivial if $\|T^*k\| = 0$.

Therefore $T^*$ is bounded. $\square$

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Does anyone know if my 3 solutions are correct? ...anyone?