Is $e^{ax}$ a bounded operator in $\mathcal{L}^2_{[0,1]}$?

Question

A similar question was asked a couple years ago:

Is multiplication with $e^{ax}$ a bounded operator in $L^2(\mathbb{R})$

But I'd like to prove it in the closed interval $[0,1]$. Suppose the operator

$\hat{A} \psi (x) = e^{ax} \psi (x)$. Computing the norm:

$||\hat{A} \psi||^2 = \int_0^1 e^{2ax} |\psi|^2 dx$, and I don't know how to continue. I've tried using the Cauchy-Schwarz inequality, as well as trying to prove its continuity, but I'm stuck. Some help?

Answer 1

It seems to me that $e^{2ax}$ is continuous on the compact $[0,1]$, so your norm is bounded by $\left\|e^{2 a x}\right\|_{\infty} \int_0^1|\psi(x)|^2 d x$, which is finite because $\psi$ is in $L^2$.