Operator norm of $Tf(x)=\int_0^x tf(t)dt$

Question

I am trying to prove that the operator norm for $$Tf(x)=\int_0^x tf(t)dt$$ with $T:C[0,a]\rightarrow C[0,a]$ and norm $\|f\|_a=\max_{x\in[0,a]}|f(x)|e^{-ax}$ is $$\|T\|=\max_{x\in[0,a]} e^{-ax}\int_0^x te^{at}dt.$$

For the lower bound, I've taken $f(t)=e^{at}$ which gives me $$\|T\|\geq \max_{x\in[0,a]} e^{-ax}\int_0^x te^{at}dt.$$

I don't know how to do the upper bound part.

Answer 1

By definition of $\|f\|_a$, we have $$|f(x)| e^{-ax} \le \| f\|_ a\Rightarrow |f(x)|\le \| f\|_a e^{ax}.$$

So

$$|Tf (x) |= \left| \int_0^x t f(t)dt\right| \le \int_0^x t |f(t)| dt \le \| f\|_a \int_0^x t e^{at} dt .$$

So

$$ \|Tf\|_a = \sup_{x\in [0,a]} e^{-ax} |Tf(x)| \le \|f\|_a \left(\sup_{x\in [0,a]} e^{-ax} \int_0^x t e^{at} dt \right) $$