Boundedness of Volterra operator with Sobolev norm Consider the subspace of $C^\infty([0,1])$ functions in the Sobolev space $H^1$.  I want to know whether the Volterra operator
\begin{equation}
V(f)(t) = \int_0^t f(s) \, ds
\end{equation}
is bounded as a linear operator from $(C^\infty([0,1]), \lVert \cdot \rVert_{1,2})$ to itself.  To be clear, the norm I'm using is
\begin{equation}
\lVert f \rVert_{1,2} = \left( \int_0^1 f^2 + (\frac{df}{dx})^2 \, dx \right)^{1/2}.
\end{equation} 
I'm having trouble bounding the value of the function by its derivative, and would like some help with this or an example to show that $V$ is not bounded.  
 A: Since $(V(f))' = f$, it suffices to see that $\lVert V(f)\rVert_{L^2} \leqslant C\lVert f\rVert_{1,2}$. But that is a direct consequence of the continuity of the Volterra operator on $L^2([0,1])$,
$$\begin{align}
\int_0^1 \lvert V(f)(t)\rvert^2\,dt &=\int_0^1\left\lvert \int_0^t f(s)\,ds\right\rvert^2\,dt\\
&\leqslant \int_0^1 \left( \int_0^t \lvert f(s)\rvert\,ds\right)^2\,dt\\
&\leqslant \int_0^1 \left(\int_0^t 1^2\,ds\right)\left(\int_0^t \lvert f(s)\rvert^2\,ds\right)\,dt\\
&\leqslant \int_0^1 t\lVert f\rVert_{L^2}^2\,dt\\
&= \frac{1}{2}\lVert f\rVert_{L^2}^2.
\end{align}$$
Hence we have
$$\lVert V(f)\rVert_{1,2}^2 \leqslant \frac{3}{2} \lVert f\rVert_{L^2}^2,$$
and we see that the Volterra operator is even continuous from $L^2([0,1])$ to $H^1$, thus a foritori as an operator $H^1\to H^1$.
A: The estimate
$$
|Vf(t)|
\leq
\int_0^1|f|
\leq
\left(\int_0^1|f|^2\right)^{1/2}
$$
gives $\|Vf\|_{L^2}\leq\|f\|_{L^2}$.
Since $(Vf)'=f$, this gives
$$
\|Vf\|_{1,2}^2
=
\|Vf\|_{L^2}^2
+
\|(Vf)'\|_{L^2}^2
\leq
2\|f\|_{L^2}^2
\leq
2\|f\|_{1,2}^2.
$$
This gives continuity $V:H^1\to H^1$ (and $L^2\to H^1$).
If you have trouble bounding the value of a function by its derivative in future, you might want to learn about the inequalities of Poincaré, Friedrichs and Sobolev.
