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Let $V:L^2[0,1]\to L^2[0,1]$ be the Volterra operator given by $f\mapsto V(f)$ where $$V(f)(t)=\int_0^tf(s)ds,\ \forall t\in[0,1].$$

My question is: Is it true that for for each $d>0$ small there exists $f\in L^2[0,1]$ such that $$\parallel V(f)-f\parallel_{L^2[0,1]}<d\ \ \text{and}\ \ \parallel f\parallel_{L^2[0,1]}\ge\sqrt{d}\ \ \ ?$$

I've tried to find functions such that $V(f)$ is similar to $f$ so that $\parallel V(f)-f\parallel_{L^2[0,1]}$ is small (such as $f(t)=e^t$ or something like that) but it didn't help in anything. Any help will be appreciated! Thank you so much.

J. W. Tanner
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Carlos
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1 Answers1

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The spectral radius of $V$ is equal $0,$ see the spoiler. Hence the operator $V-I$ is invertible. Thus there exists a constant $c>$ such that $$\|V(f)-f\|_2\ge c\|f\|_2$$ Therefore the claim in OP is not satisfied for arbitrary small values of $d.$

It can be proved by induction that $$(V^nf)(x)={1\over (n-1)!}\int\limits_0^x(x-y)^{n-1}f(y)\ dy\quad (*)$$ Therefore $$\|V^n\|\le {1\over (n-1)!}$$ The series $$\sum_{n=0}^\infty \lambda^{-n-1}V^n$$ is therefore convergent and represents the operator $(\lambda I-V)^{-1}.$ By $(*)$ we get $$[(\lambda I-V)^{-1}f](x)=\lambda^{-1}f(x)+\lambda^{-2}\int\limits_0^x e^{(x-y)/\lambda}f(y)\ dy$$ We can actually ignore the previous reasoning and verify directly that the RHS of the last formula is indeed the inverse of $\lambda I-V.$

Ryszard Szwarc
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  • Great answer. I have a followup question. Is there a direct way of proving that $V-I$ is invertible? Do we really need to do that (ingenious) detour via the spectral radius formula? Say, for example, I guess that the fact that $\ker(V-I)=0$ follows from the fact that the ODE initial value problem $$\begin{cases} f'=f \ f(0)=0\end{cases}$$ only has the trivial solution. – Giuseppe Negro May 05 '23 at 17:14
  • @GiuseppeNegro Thanks. I have modified the spoiler by giving explicitly the inverse of $\lambda I-V$ for $\lambda\neq 0.$ – Ryszard Szwarc May 05 '23 at 19:24
  • This is exactly what I was asking for. I guess you could also compute that inverse via ODE methods. The result, of course, is the same (as it should be). – Giuseppe Negro May 05 '23 at 21:04