# Question about linear operators

• AxiomOfChoice
In summary: Yes, but A is a continuous linear operator (as stated in the OP), so \{\|Ax\|~\vert~x\leq 1\}is bounded, by the boundedness of \|Ax\|.
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?

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 said:
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?

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AxiomOfChoice said:
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...

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micromass said:
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?

AxiomOfChoice said:
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!

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.

Bacle said:
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...

My bad, did not read carefully. Sorry.

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

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micromass said:
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?

AxiomOfChoice said:
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).

## 1. What are linear operators?

Linear operators are mathematical functions that map one vector space to another in a linear manner. They are used in various fields of science and engineering to model and manipulate linear systems.

## 2. How are linear operators different from other operators?

Unlike other types of operators, linear operators preserve the basic algebraic properties of vector spaces, such as linearity and scaling. This means that the output of a linear operator is always a linear combination of the inputs.

## 3. What are the applications of linear operators?

Linear operators are widely used in fields such as physics, engineering, mathematics, and computer science to model and solve linear systems. They are also used in signal processing, quantum mechanics, and control theory.

## 4. Can linear operators be represented by matrices?

Yes, linear operators can be represented by matrices. This is because matrices can efficiently perform the operations of linear operators, such as addition, multiplication, and inversion. In fact, many common linear operators, such as derivatives and integrals, have matrix representations.

## 5. What are the properties of linear operators?

Some of the key properties of linear operators include linearity, commutativity, and associativity. They also have a neutral element (identity operator) and an inverse element. Additionally, linear operators can be composed together to form new operators.

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