Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

2 Answers 2

up vote 5 down vote accepted

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 $H^1\to H^1$.

share|improve this answer

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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