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.


2 Answers 2


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 $H^1\to H^1$.


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.


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