Equivalence of several definitions of operator norm

BAYMAX

Aug 26, 2017 17:12

We were trying to prove theorem 6.6 of this text!

Martin Sleziak

It's Limaye's Funcional Analysis, if somebody wonders.

Relevant notation:

$\|F\|=\sup\{\|F(x)\|; \|x\|\le1\}$

$\alpha_0=\inf\{\alpha\ge0; (\forall x\in X) \|F(x)\|\le\alpha\|x\|\}$

$\beta=\sup\{\|F(x)\|; \|x\|=1\}$

$\gamma=\sup\{\|F(x)\|; \|x\|=1\}$

where $F\colon X\to Y$ is a linear map between two linear normed spaces.

And the point of the whole thing is to show that these four values are equal to each other: $\|F\|=\alpha=\beta=\gamma$.

And the book is Limaye: Functional Analysis. (But I guess my typo in the message above does not matter that much.)

BAYMAX

actually $\gamma = sup\{||F(x)|| : x \in X , ||x||<1\}$

Martin Sleziak

Thanks for the correction.

BAYMAX

my pleasure!

Martin Sleziak

As I mentioned before $\beta\le\|F\|$ should be immediate. And also $\gamma\le\|F\|$ can be seen easily.

Both of them follow simply from this fact: $A\subseteq B$ $\Rightarrow$ $\sup A\le \sup B$.

BAYMAX

Aug 26, 2017 17:19

nice

Martin Sleziak

Perhaps it is worth mentioning that we know that the suprema in the definition of $\|F\|$, $\beta$, $\gamma$ are finite, since it we assume that $F$ is continuous. (I have forgotten to mention this assumption above.)

Anyway, probably proof that all four values are equal to $\infty$ if $F$ is not continuous would be almost the same.

BAYMAX

yes

Martin Sleziak

Ok, so we have at least some inequalities.

Another one which, I think, is relatively easy is $\|F\|\le\beta$.

We only need to show that $\|F(x)\|\le\beta$ for each $\|x\|\le1$.

Do you have some idea how to do this?

BAYMAX

is that from $||F|| \leq \beta ||x||$

as we are given $||x|| \leq 1$

Martin Sleziak

Ok, that's a different from what I had in mind, but we might try this way.

How do you get that $\|F\| \le \beta \|x\|$?

Did you mean to write $\|F(x)\| \le \beta\|x\|$?

BAYMAX

Aug 26, 2017 17:28

I see my error

Martin Sleziak

What about this.

We consider some point $x$ such that $\|x\|\le1$.

BAYMAX

then $||F(x)|| \leq \beta||x|| \leq \beta$

Martin Sleziak

Can we find $\|y\|$ such that $\|y\|=1$ and $\|F(x)\| \le \|F(y)\|$?

@BAYMAX Ok, if we want to try it this way, how do we know $\|F(x)\| \le \beta \|x\|$?

BAYMAX

as $F(x) \in BL(X,Y)$

Martin Sleziak

Yes, I see that $F(x)\in BL(X,Y)$ implies that both $\|F\|$ and $\beta$ are finite.

But how do you get the inequality $\|F(x)\| \le \beta \|x\|$?

I'll let you talk for a bit, sorry if I cut you off mid-proof.

BAYMAX

Aug 26, 2017 17:32

$F(x)$ is a bounded linear map

so we have from the definition $||F(x)|| \leq \beta ||x||$

that is there exists some $\beta > 0$ such that $||F(x)|| \leq \beta ||x||$

Martin Sleziak

This is very bad notation.

You're using $\beta$ for two different things.

Just a few lines above you defined $\beta=\sup\{\|F(x)\|; \|x\|=1\}$.

And now you are using the letter $\beta$ for one of the number which exist by the definition of boundedness.

You do not know whether those "two betas" are in fact the same.

BAYMAX

hmm..sorry..i must have thought that this $\beta$ may not work for the case of bounded linear map..

Martin Sleziak

Anyway, let's try get back to this:

Let's assume that we have an $x$ with $\|x\|\le 1$.

BAYMAX

I am messing with notations as I see

ok

Martin Sleziak

Can we find $\|y\|$ such that $\|y\|=1$ and $\|F(x)\| \le \|F(y)\|$?

Just think of this geometrically. You have a point inside unit ball. Can you somehow find a "corresponding" point on the boundary of the unit ball?

BAYMAX

Aug 26, 2017 17:39

may be silly but i am thinking of extending the point inside the unit ball to the boundary of the unit ball?

Martin Sleziak

That's exactly what we want to do.

BAYMAX

oh

Martin Sleziak

So if we have a point $x$ and we want to multiply this by constant $c>0$ in such way that the result will be on the boundary.

BAYMAX

yes!

Martin Sleziak

I.e., we want $y=cx$ and we want to choose $c$ in such way that $c>0$ and $\|y\|=\|cx\|=1$.

BAYMAX

Aug 26, 2017 17:40

yes

Martin Sleziak

It's not difficult to see that the only possible choice is $c=\frac1{\|x\|}$.

BAYMAX

yes $c = \frac{1}{||x||}$

Martin Sleziak

And it's also easy to see that $c\ge1$ (since $\|x\|\le1$).

So far ok?

BAYMAX

yes,so far nice!

Martin Sleziak

Now we get $F(y)=F(cx) = c F(x) \ge F(x)$.

Which means $$F(x)\le F(y).$$

But we also have $\|y\|=1$ and thus $$F(y)\le\beta$$ directly from the definition of $\beta$.

And now just by putting these two inequalities together we get $$F(x)\le F(y)\le\beta.$$

Is this ok?

BAYMAX

Aug 26, 2017 17:43

yup

Martin Sleziak

As I mentioned, it might help to draw a picture. (But it is difficult to draw a picture here in chat.)

Anyway to summarize, so far we have that $$\|F(x)\|\le\beta\tag{1}$$ for every $x\in X$ such that $\|x\|\le1$.

But this implies that $$\sup\{\|F(x)\|; \|x\|\le 1\}\le\beta.\tag{2}$$

But $(2)$ is nothing else, just $\|F\|\le\beta$. So we have both inequalities and we have finally shown $\boxed{\|F\|=\beta}$.

BAYMAX

I think in the above

$||F(y)|| \leq \beta$

that is $||y|| = 1$ implying $||F(y)|| \leq \beta$

?

Martin Sleziak

Yes, this is from definition of $\beta$.

BAYMAX

5 mins ago, by Martin Sleziak

But we also have $\|y\|=1$ and thus $$F(y)\le\beta$$ directly from the definition of $\beta$.

Martin Sleziak

And I appologize, I should have norms in what I wrote above.

BAYMAX

Aug 26, 2017 17:48

no problem i was just confirming

Martin Sleziak

The message you quoted should have been $$\|F(y)\|\le\beta.$$

And similarly I've forgotten norms in $$\|F(x)\|\le\|F(y)\|.$$

Which I get from $\|F(y)\|=\|F(cx)\|=c\|F(x)\|\ge \|F(x)\|$.

The last inequality follows from $c\ge 1$ and $\|F(x)\|\ge0$.

So with there corrections, the above should be proof of $\boxed{\|F\|=\beta}$.

BAYMAX

yes so we got $||F|| = \beta$!

Martin Sleziak

Ok, so that's one part of theorem.

Are we going to try $\|F\|=\gamma$ next?

BAYMAX

Let's do it!

Martin Sleziak

Again, we already know $\gamma\le\|F\|$.

So proving $\beta\le\gamma$ or $\|F\|\le\gamma$ would be enough.

BAYMAX

Aug 26, 2017 17:52

yup, next we try to show $||F|| \leq \gamma$

Martin Sleziak

Suppose we have $x$ such that $\|x\|\le 1$.

BAYMAX

yes

Martin Sleziak

If is easy to find a sequence $t_n$ such that $t_n\to x$ and $\|t_n\|<1$.

For example, we could take $t_n=\left(1-\frac1n\right)x$.

BAYMAX

yes

also $||t_{n}|| < 1$

Martin Sleziak

This also implies $F(t_n) \to F(x)$ since $F$ is continuous.

And (this is an important step) we also get $\|F(t_n)\| \to \|F(x)\|$, since the norm is continuous.

BAYMAX

Aug 26, 2017 17:56

Hmm..yes

Martin Sleziak

We are using the fact that $x\mapsto\|x\|$ is a continuous function. Perhaps you have derived this before.

BAYMAX

I will check it back! yes $x \rightarrow ||x||$ is a continuous function

so $||F(t_{n})|| \rightarrow ||F(x)||$

Martin Sleziak

I guess there are even a few post on main about proving that norm is continuous (in a normed space) or that distance is continuous (in a metric space).

Ok, so we have $\|F(x)\|=\lim\limits_{n\to\infty} \|F(t_n)\|$.

BAYMAX

yes

Martin Sleziak

At the same time we know that $\|F(t_n)\|\le\gamma$ for each $n$.

BAYMAX

Aug 26, 2017 18:01

yes

Martin Sleziak

Together we get that $$\|F(x)\|\le\gamma.\tag{*}$$

Just from the fact that limits preserve (non-strict) inequalities.

BAYMAX

nice

Martin Sleziak

Now since $(*)$ is true for any $\|x\|\le1$, we get $$\sup\{\|F(x)\|; \|x\|\le1\}\le\gamma.$$

But this is exactly $\|F\|\le\gamma$.

So now we have $\boxed{\|F\|=\gamma}$.

BAYMAX

Gotcha :)

Martin Sleziak

So the only question that remains is what to do about $\alpha_0$.

BAYMAX

Aug 26, 2017 18:04

ya we next try to prove $||F|| = \alpha_{0} $

Martin Sleziak

Perhaps let me first try to prove that for every $t\in X$ we have $$\|F(t)\|\le\|F\|\cdot\|t\|.\tag{A}$$

BAYMAX

Oh yes it was there!

Yes

Martin Sleziak

Or if you prefer we can discuss how $(A)$ implies $\alpha_0 \le \|F\|$ to see that this is in fact related to $\alpha_0$.

Whichever you prefer.

So do we show $(A)$ first and then show that it gives us one of the inequalities later?

BAYMAX

Let's discuss (A)

Martin Sleziak

Ok.

So know we have $t\in X$.

This $t$ can be arbitrary.

BAYMAX

Aug 26, 2017 18:08

yes

Martin Sleziak

We want to say something about $t$ using $\|F\|$, but $\|F\|$ is defined only using values for $\|x\|\le1$.

So again, as before, we are going to scale.

BAYMAX

yes divide by $||t||$

right?

Martin Sleziak

Let us define $$x=\frac t{\|t\|}.$$

Exactly.

BAYMAX

nice!

Martin Sleziak

We should be able to see that $\|x\|=1$.

BAYMAX

Aug 26, 2017 18:09

yes

$||x|| =1$

Martin Sleziak

And this implies $\|F(x)\|\le\|F\|$.

What does this say about $t$ and especially about $\|F(t)\|$?

We have $$\|F(t)\| = \|F(\|t\|\cdot x) = \|t\|\cdot \|F(x)\| \le \|t\| \cdot \|F\|.$$

That is we have shown that $$\|F(t)\| \le \|F\|\cdot\|t\|$$ holds for any $t\in X$.

So we have proved $(A)$.

I'll wait a bit to see whether you are ok with the proof of $(A)$.

BAYMAX

sorry my system chrome hanged a bit

Martin Sleziak

Again, you can also try to draw a picture. We used again the same thing - we tried to scale the given vector.

BAYMAX

Yes

Its ok with proof (A)

Martin Sleziak

Ok.

And I want to show that if we have $(A)$ then we can get $\alpha_0\le\|F\|$ from it.

Recall that $\alpha_0=\inf\{\alpha\ge0; (\forall x\in X) \|F(x)\|\le\alpha\|x\|\}$.

In fact, to simplify things let us denote $S=\{\alpha\ge0; (\forall x\in X) \|F(x)\|\le\alpha\|x\|\}$.

Then we have $\alpha_0=\inf S$.

BAYMAX

Aug 26, 2017 18:17

Yes

that's good!

Martin Sleziak

Now it suffices to notice that from $(A)$ we see that $$\|F\|\in S.$$

And this, clearly, implies $\inf S \le \|F\|$.

BAYMAX

yes $||F|| \in S$

yes

$\alpha_{0} \leq ||F||$

Martin Sleziak

Ok so we have $\boxed{\alpha_0\le\|F\|}$

BAYMAX

yup

Martin Sleziak

The only one missing is $\|F\|\le\alpha_0$.

BAYMAX

Aug 26, 2017 18:20

yes

Martin Sleziak

Or, equivalently, $\beta\le\alpha_0$ would be sufficient, too.

BAYMAX

yes

Martin Sleziak

We will still use the notation $S=\{\alpha\ge0; (\forall x\in X) \|F(x)\|\le\alpha\|x\|\}$.

Let us take arbitrary $x$ such that $\|x\|=1$.

BAYMAX

yes sure.

Martin Sleziak

I am looking at points at the boundary, since such points appear in the definition of $\beta$.

Let us consider arbitrary $\alpha\in S$.

Then we immediately see $\|F(x)\|\le\alpha$.

BAYMAX

Aug 26, 2017 18:22

ok

yes

Martin Sleziak

Since since this is true for every $\|x\|=1$ and $\beta=\sup\{\|F(x)\|; \|x\|=1\}$, we also get $$\beta\le\alpha.$$

(Just using the definition of supremum.)

BAYMAX

yes

Martin Sleziak

And once again the same trick - but this time we will use infimum.

The above is true for every $\alpha\in S$.

So we get $\beta\le\inf S$.

BAYMAX

yes this must be true for the $inf (S)$

Martin Sleziak

Which is exactly $\beta\le\alpha_0$.

BAYMAX

Aug 26, 2017 18:24

yes $\beta \leq inf S$

yup

so $\beta = \alpha_{0}$

Martin Sleziak

And the proof is finished.

BAYMAX

so $\beta = \alpha_{0} = \gamma = ||F||$

yes the proof is finished :)

$Mathematics$ Modern Abstract Analysis

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