# Calculating the norm of linear, continuous operator

1. Nov 22, 2013

### mahler1

The problem statement, all variables and given/known data.

Let $X=\{f \in C[0,1]: f(1)=0\}$ with the $\|x\|_{\infty}$ norm. Let $\phi \in X$ and let $T_{\phi}:X \to X$ given by

$T_{\phi}f(x)=f(x)\phi(x)$.
Prove that $T$ is a linear continuous operator and calculate its norm.

The attempt at a solution.

To check for linearity is pretty simple, for continuity I am not so sure if my proof is correct so I write it just in case:

$T$ is continuous if and only if $T$ is bounded. So, I want to prove that there exists $c>0: \forall f \in X$, $\|T(f)\|_{\infty}\leq c\|f\|_{\infty}$. $\|T(f)\|_{\infty}=sup_{x \in [0,1]}|f(x)\phi(x)|\leq sup_{x \in [0,1]}|f(x)|sup_{x \in [0,1]}|\phi(x)|$. As $[0,1]$ is compact, $sup_{x \in [0,1]}|\phi(x)|=max_{x \in [0,1]}|\phi(x)|$, if I call $c=max_{x \in [0,1]}|\phi(x)|$, then $sup_{x \in [0,1]}|f(x)\phi(x)|\leq csup_{x \in [0,1]}|f(x)|$ for every $f \in X$. It follows that $T$ is bounded $\implies$ $T$ is continuous.

Now, I had problems to calculate the norm of this operator. One of the definition for the norm in the space of linear, continuous operators is $||T||=inf\{M: \|T(f)\|_{\infty}\leq M\|f\|_{\infty}\}$ (among other equivalent definitions). Sorry if I say something trivially false but I think that $c=max_{x \in [0,1]}|\phi(x)|$ is the norm. I have to prove that $c$ is a lower bound of the set $\{M: \|T(f)\|_{\infty}\leq M\|f\|_{\infty}\}$ and that it is the upper lower bound.

Here's a tentative solutions after the suggestions:
Lets show that $c=max_{x \in [0,1]}|\phi(x)|$ is a lower bound of the set $A=\{M: \|T(f)\|_{\infty}\leq M\|f\|_{\infty}\}$. Let $M \in A$, then $\|T(\phi)\|_{\infty}\leq M \|\phi\|_{\infty}$. So $c^2=sup_{x \in [0,1]}|\phi(x)\phi(x)| \leq Mc \implies c\leq M$. This means that $c$ is a lower bound of $A$, but $sup_{x \in [0,1]}|f(x)\phi(x)|\leq csup_{x \in [0,1]}|f(x)|$ for every $f \in X$, so $c \in A$. It follows that $||T||=inf\{M: \|T(f)\|_{\infty}\leq M\|f\|_{\infty}\}=c$.

Last edited: Nov 22, 2013
2. Nov 22, 2013

### LCKurtz

Look at the norm of $T_\phi (\phi)$.

3. Nov 22, 2013

### Dick

I think your proof has a lot of the right ideas in it. But it's really unclear. And it does contain false statements. $sup |f \phi| \le sup |f| sup |\phi|$. They aren't equal. And X isn't compact. [0,1] is compact. And LCKurtz has a great hint. And I don't think f(1)=0 has much to do with this part of the exercise. Can you rearrange the whole thing and try again?

4. Nov 22, 2013

### mahler1

It really helped that hint, thanks!

5. Nov 22, 2013

### mahler1

I meant $[0,1]$, no idea why I wrote $X$. If the elements of two sets $A$ and $B$ consist of non-negative elements, are you sure that $sup AB \leq sup A sup B$ and not always equal? I'll think about a counterexample to convince myself. In the last part of the exercise I've used that if the elements of $A$ are non-negative numbers, then $supAA=supAsupA$.

6. Nov 23, 2013

### Dick

I'm absolutely sure. Try f(x)=1-x and g(x)=(1-x)*x. And you are dealing with sup|f|, it doesn't much matter whether anything is nonnegative or not.