I have looked at the proofs in :
Norm of integral operator in $L_2$
and Halmos' solution to problem 188 but i wanted to find a more relaxed bound of the volterra operator without using the Hilbert adjoint operator. $$(Vx)(t)=\int_0^tx(s)ds$$ $ V : L^2(0,1)\to L^2(0,1)$, $t \in [0,1]$, $x \in L^2(0,1)$ .
I need to use the Cauchy Shwartz Inequality to prove that $||V||\leq \frac{1}{\sqrt{2}}$ .
I have already tried finding the norm using the following equation$$||||=\frac{||||}{||||}$$ and with
$$||||=sup||||$$ with $||x||=1$ using the $L_2$ norm. But these solutions tend to provide little insight without any manipulations.
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Jack
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Where have you tried to use Cauchy-Schwarz? – Severin Schraven Dec 08 '19 at 19:25
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@SeverinSchraven I tried using it in the inner product definition of cauchy shwartz. I did not think about the integral cauchy shwartz equivalent – Jack Dec 08 '19 at 20:31
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1Those are actually the same, as we can take the $L^2$ inner product $\langle f, g \rangle = \int_0^1 f(x) g(x) dx$. – Severin Schraven Dec 08 '19 at 20:32
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By Cauchy-Schwarz we have
$$ \vert (Vx)(t) \vert =\left\vert \int_0^t x(s) ds \right\vert \leq \int_0^t 1\cdot \vert x(s) \vert ds \leq \left( \int_0^t 1 ds \right)^\frac{1}{2} \left( \int_0^t \vert x(s) \vert^2 ds \right)^\frac{1}{2} \leq \sqrt{t} \Vert x \Vert_{L^2(0,1)}$$ Therefore, we get $$ \Vert Vx \Vert_{L^2(0,1)} = \left( \int_0^1 \vert (Vx)(t) \vert^2 dt \right)^\frac{1}{2} \leq \left( \int_0^1 t \ dt \right)^\frac{1}{2} \Vert x \Vert_{L^2(0,1)} = \frac{1}{\sqrt{2}} \Vert x \Vert_{L^2(0,1)}$$ Hence, we get $$ \Vert V \Vert \leq \frac{1}{\sqrt{2}} $$
Severin Schraven
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