# Functional Analysis problems need checking

1. Jul 28, 2005

### Oxymoron

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 $\mathbb{R}$.

2. Jul 28, 2005

### Oxymoron

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

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

Last edited: Jul 28, 2005
3. Jul 28, 2005

### Galileo

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.

4. Jul 29, 2005

### Oxymoron

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

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

5. Jul 29, 2005

### Oxymoron

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 $\{a_n\} \in l^{\infty}$.

Last edited: Jul 29, 2005
6. Jul 29, 2005

### 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.

7. Jul 29, 2005

### Oxymoron

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.

Last edited: Jul 29, 2005
8. Jul 29, 2005

### MalleusScientiarum

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

9. Aug 1, 2005

### Oxymoron

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

10. Aug 4, 2005

### Oxymoron

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.

11. Aug 4, 2005

### Oxymoron

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 $T^*$. $\square$

Last edited: Aug 4, 2005
12. Aug 4, 2005

### Oxymoron

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. $\square$

Last edited: Aug 5, 2005
13. Aug 5, 2005

### Oxymoron

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. $\square$

Last edited: Aug 5, 2005
14. Aug 8, 2005

### Oxymoron

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