Show that the operator $T:L^2[0,1]\to L^2[0,1]$ is bounded and find its norm, where $Tf(t)=\int_0^t e^{-(t-x)}f(x)dx$.
My calculation:\begin{aligned} \|Tf\|^2&=\int_0^1|\int_0^t e^{-(t-x)}f(x)dx|^2dt\\ &\leq \int_0^1(\int_0^t e^{-(t-x)}dx)(\int_0^t|f(x)|^2dx)dt\\ &=\int_0^1(1-e^{-t})(\int_0^t|f(x)|^2dx)dt\\ &\leq (1-e^{-1})\int_0^1(\int_0^t|f(x)|^2dx)dt\\ &=(1-e^{-1})\|f\|^2 \end{aligned} Then $\|Tf\|\leq\sqrt{(1-e^{-1})}\|f\|$ and the norm is $\|T\|=\sup_{\|f\|=1}\|Tf\|=\sqrt{(1-e^{-1})}$.
I don't know whether my calculation is correct. Could someone take a look? Thanks!
Assuming you're using Cauchy-Schwarz, your second line should be $$\int_0^1(\int_0^t e^{-2(t-x)}dx)(\int_0^t|f(x)|^2dx)dt$$
Also, your last line should be an inequality, not an equality, as in $$(1-e^{-1})\int_0^1\left(\int_0^t|f(x)|^2\,dx\right)\,dt \leq(1-e^{-1})\|f\|^2$$
Last, if it all were correct, we would only be able to conclude $\|T\| \leq \sqrt{1-e^{-1}},$ not an equality.