Find the norm of the linear operator $Ax=t^2x(t)$

Question

How do I find the norm of this linear operator : $Ax=t^2x(t)$ where $x(t)$ is from $C[a,b]$.

I tried to show that $A$ is continuous. Here what I did:

let's take $x_0(t)$ from $C[a,b]$.

For any $\epsilon>0$ there is $\delta>0$ that $||x(t)-x_0(t)||<\delta$.

$||Ax-Ax_0||= ||t^2x(t)-t^2x_0(t)||\leq\ t^2||x(t)-x_0(t)||<t^2\delta<\epsilon $

if we take $\delta<\epsilon/t^2$ Operator $A$ will be continuous, so It'll be bounded.

To find the norm I took $x\ne0$ from $C[a,b]$:

$||A||=sup||Ax||/||x|| = sup||t^2(x/||x||)|| = b^2$

Is this correct proof? Am I right?

Answer 1

Your answer is correct, but not the reasoning. The norm on $C[a,b]$ is $\|x\|=\max_{a\le t\le b}|x(t)|$. Then $$ \|A\,x\|=\max_{a\le t\le b}|t^2\,x(t)|\le b^2\,\|x\|. $$ This shows that $\|A\|\le b^2$. To show that $\|A\|=b^2$ pick the constant function $x(t)=1$. Then $\|x\|=1$ and $\|A\,x\|=\max_{a\le t\le b}t^2=b^2$.