Question about linear operators

[AxiomOfChoice]

Apparently - that is, if I'm to believe Kolmogorov - we have the following for a bounded linear operator A between two normed spaces:

$$\sup_{\| x \| \leq 1} \|Ax\| = \sup_{\|x\| = 1} \|Ax\|$$

But why?

 

[micromass]

Hi AxiomOfChoice!

First of all, it is obvious that

$$\sup_{\|x\|=1}{\|Ax\|}\leq \sup_{\|x\|\leq 1}{\|Ax\|}$$

since you take the supremum over more. For the converse, say that we have a sequence $x_n$ with $\|x_n\|\leq 1$ such that $\|Ax_n\|\rightarrow \sup_{\|x\|\leq 1}{\|Ax\|}$. Then put

$$y_n=\frac{x_n}{\|x_n\|}$$

Then $\|y_n\|=1$, and

$$\|Ax_n\|\leq\|Ay_n\|$$

thus $\sup\|Ax_n\|\leq \sup\|Ay_n\|$...

 

[AxiomOfChoice]

Hi AxiomOfChoice!

First of all, it is obvious that

$$\sup_{\|x\|=1}{\|Ax\|}\leq \sup_{\|x\|\leq 1}{\|Ax\|}$$

since you take the supremum over more. For the converse, say that we have a sequence $x_n$ with $\|x_n\|\leq 1$ such that $\|Ax_n\|\rightarrow \sup_{\|x\|\leq 1}{\|Ax\|}$. Then put

$$y_n=\frac{x_n}{\|x_n\|}$$

Then $\|y_n\|=1$, and

$$\|Ax_n\|\leq\|Ay_n\|$$

thus $\sup\|Ax_n\|\leq \sup\|Ay_n\|$...

micromass: Thanks so much for your quick response! I actually posted about something related to this several months ago, and you were kind enough to offer your assistance then, too! So, thanks once again.

However, in my original post about this, which can be found https://www.physicsforums.com/showthread.php?p=3184217#post3184217", someone answered the question YOU just answered very succinctly: "a linear functional will attain its maximum on the boundary of the set." Why is THAT true?

 

[micromass]

micromass: Thanks so much for your quick response! I actually posted about something related to this several months ago, and you were kind enough to offer your assistance then, too! So, thanks once again.

However, in my original post about this, which can be found https://www.physicsforums.com/showthread.php?p=3184217#post3184217", someone answered the question YOU just answered very succinctly: "a linear functional will attain its maximum on the boundary of the set." Why is THAT true?

The easiest way to see this is by using the open mapping theorem. Consider a continuous nonzero functional A on a Banach space. If K is compact then A(K) will attain a maximum value. However, let U be the interior of K. By the open mapping theorem, we see that A(U) must be open. So A(U) does not have a maximum. So the maximum of A(K) can not be attained on the interior, but it must lie on the boundary...

 

[AxiomOfChoice]

The easiest way to see this is by using the open mapping theorem. Consider a continuous nonzero functional A on a Banach space. If K is compact then A(K) will attain a maximum value. However, let U be the interior of K. By the open mapping theorem, we see that A(U) must be open. So A(U) does not have a maximum. So the maximum of A(K) can not be attained on the interior, but it must lie on the boundary...

I humbly ask that you walk me through your argument a bit more carefully All I know of the open mapping theorem is that a continuous linear map from one Banach space onto another Banach space is an open mapping. Do we know that our functional - which I'm guessing is $f(x) = \|Ax\|$ so that $f: X \to \mathbb R$ - is onto?

 

[micromass]

I humbly ask that you walk me through your argument a bit more carefully All I know of the open mapping theorem is that a continuous linear map from one Banach space onto another Banach space is an open mapping. Do we know that our functional - which I'm guessing is $f(x) = \|Ax\|$ so that $f: X \to \mathbb R$ - is onto?

Ah, it seems that I was too fast with my argument. Indeed, my functional would be $\|Ax\|$, but this isn't linear. So I'll have to be a bit more careful.

Anyway, A is a linear operator which is surjective on it's image. So take U open, then A(U) is an open set of it's image. Since taking the norm $\|~\|$ is an open mapping, we knowing that $\|A(U)\|$ is an open set, and thus not has a maximum. The only problem that could arise is that A is contant 0, thus Ax=0 everywhere. Then A(U)={0} and then $\|A(U)\|=0$ is not open and does have a maximum. However, this can only happen if A=0, and then it is easily seen that A also attains it's maximum on the boundary!

 

[Bacle]

But, Micromass:

Don't you also need ||Ax|| to be bounded for it to be continuous? AFAIK, for linear operators, bounded is equivalent to continuous.

 

[micromass]

But, Micromass:

Don't you also need ||Ax|| to be bounded for it to be continuous? AFAIK, for linear operators, bounded is equivalent to continuous.

Yes, but A is a continuous linear operator (as stated in the OP), so

$$\{\|Ax\|~\vert~x\leq 1\}$$

is bounded, by definition...

 

[Bacle]

My bad, did not read carefully. Sorry.

Another tricky one is ||Ax||<=||A||||x||

 

[AxiomOfChoice]

Ah, it seems that I was too fast with my argument. Indeed, my functional would be $\|Ax\|$, but this isn't linear. So I'll have to be a bit more careful.

Anyway, A is a linear operator which is surjective on it's image. So take U open, then A(U) is an open set of it's image. Since taking the norm $\|~\|$ is an open mapping, we knowing that $\|A(U)\|$ is an open set, and thus not has a maximum. The only problem that could arise is that A is contant 0, thus Ax=0 everywhere. Then A(U)={0} and then $\|A(U)\|=0$ is not open and does have a maximum. However, this can only happen if A=0, and then it is easily seen that A also attains it's maximum on the boundary!

Okay! And the fact that the norm is an open mapping follows from an application of $| \|x\| - \|y\|| \leq \|x - y\|$, right? Or is there an easier/more elegant way to see this?

 

[micromass]

Okay! And the fact that the norm is an open mapping follows from an application of $| \|x\| - \|y\|| \leq \|x - y\|$, right? Or is there an easier/more elegant way to see this?

Yes, that's right!
Another way to see it is that the image of open balls is open. For example the image of $B(0,\varepsilon)$ is $[0,\epsilon[$, which is open. (in the positive reals of course).