3
$\begingroup$

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.

$\endgroup$
6
$\begingroup$

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 $H^1\to H^1$.

$\endgroup$
4
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.