Let Hilbert space $H=L^2((0,\infty),dt)$ where $dt$ is the Lebesgue measure. For each $f\in H$, define $Tf:(0,\infty)\to \mathbb{C}$ by $$ (Tf)(s)=\frac{1}{s}\int_{(0,s)}f(t)dt. $$ Show that $Tf$ well-defined. What I think is, I should check if the integral converges (absolutely). Since the limit is unique in general, we must have that $Tf$ is uniquely determined, and hence it is well-defined.
Show that $Tf$ is continuous. I think it is easier to show that $Tf(s+\delta)\to Tf(s)$ as $\delta\to 0$. Then $$ \frac{1}{s+\delta}\int_{(0,s+\delta)}f(t)dt=\frac{s}{s+\delta}\left ( Tf(s)+\int_{(s,s+\delta)}f(t)dt \right ). $$ Taking $\delta \to 0$, may I use/write on the last integral to $\int_{\{s\}}f(t)dt$, which gives 0?
If $f\in L^2\bigl((0,+\infty)\bigr)$ and $s>0$, $f|_{(0,s)}\in L^2\bigl((0,s)\bigr)$. Since $1\in L^2\bigl((0,s)\bigr)$,$$f|_{(0,s)}=f|_{(0,s)}\times1\in L^1\bigl((0,s)\bigr).$$Therefore, $T$ is well defined.
Concerning continuity, you must tell us first to which space $Tf$ belongs and which topology is being used there.