What is the operator norm of this operator?

275 Views Asked by At

Let $\mathbb{X} = C^1([0,1], \mathbb{R})$ and $\|\,f\|_{\mathbb{X}} = \|\,f\|_{\infty} + \|\,f'\|_{\infty}$. Let $A: \mathbb{X} \rightarrow \mathbb{R}$ with $Af = f'\left(\frac{1}{2}\right)$.
Find the operator norm of $A$

I am not too sure how to begin here. I have several definitions of the operator norm, the one I have been trying is that $$\|A\|_{op} = \inf\big\{c \geq 0: \|Av\| \leq c \|v\|\big\}$$

But I'm not even sure what the expression for $\|Av\|$ will be...

Any help is appreciated

1

There are 1 best solutions below

8
On BEST ANSWER

Hints: Another way to look at the operator norm is to see what the operator does to a unit vector: $$ \|A\|_{op}=\sup\{|Af|: \|f\|_{\Bbb X}=1\} $$ Think about how large $|Af|$ could possibly be if we require that $\|f\|_{\Bbb X}=1$.

Some more pointers: $A$ takes in a function and spits out a real number. Which real number it spits out depends on the function, of course. For instance, $f(x)=x^2$ gives $Af=1$, and $g(x)=\sin(x)$ gives $Ag=\cos(0.5)$.

But if $f$ is a function such that $|Af|=1$, what can you say about $\|f\|_{\Bbb X}$? How small could it be? (This specific hint is more directed at the operator norm definition given in the question.)

Finally, since you were wondering, the expression for $|Av|$ is $|v'(0.5)|$.