Functional Analysis problems need checking

[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}.

 

[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

 

[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.

 

[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|

 

[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}.

 

[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.

 

[Oxymoron]

Thankyou for replying MS.

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 I am used to seeing an inequality instead of an equality - which is alarming.

 

[MalleusScientiarum]

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

 

[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?

 

[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<img src="/styles/physicsforums/xenforo/smilies/arghh.png" class="smilie" loading="lazy" alt=":H" title="Gah! :H" data-shortname=":H" />\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.

 

[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

 

[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 &amp;=&amp; \langle Th,k_1\rangle_K + \overline{\lambda}\langle Th, k_2 \rangle_K \\<br /> &amp;=&amp; \langle h,T^*k_1\rangle_K + \overline{\lambda}\langle h,T^*k_2\rangle_K \\<br /> &amp;=&amp; \langle h, T^*k_1 + \lambda T^*k_2\rangle_K <br />

Hence

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

T^* is linear. \square

 

[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\| &gt; 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

 

[Oxymoron]

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