Let $T:(C[0,1], \|\cdot\|_2)\to (C[0,1], \|\cdot\|_2)$ be defined by
$$(Tf) (s)=\int_0^1e^{-sx}f(x)\mathrm dx, s\in[0,1].$$
Show that $T$ is a bounded operator. Is $T$ one-one? Is $T$ onto? Justify.
I tried to follow this approach but I know none of the terms mentioned in it.
$|Tf(s)| \leq (\int_0^{1}|f(x)|dx)\leq (\int_0^{1}|f(x)|^{2}dx)^{1/2}=\|f\|$ and hence $\|Tf\| \leq \|f\|$, $T$ is bounded with $\|T\|\leq 1 $.
$T$ is one-to-one: If $Tf=0$ then $\int_0^{1}e^{-sx}f(x)dx=0$ for all $s \in [0,1]$. Repeatedly differentiating and putting $s=0$ we get $\int_0^{1}x^{n}f(x)dxdx=0$ for all $n \geq 0$. [At each step you have to apply DCT to justify differentiating inside the integral sign]. A standard application of Weierstrass Approximation Thoerem now show that $f=0$, so $T$ is one-to-one.
$T$ is not onto. It is easy to see $Tf$ is differentiable so non-differentiable continuous are not in the range.