Suppose $T$ is an operator on $C([0,1])$ defined by $(Tu)(t) = \displaystyle\int_{0}^{t} u(x)^2\,\mathrm dx$. Show that $T$ is a contraction mapping on the closed ball of radius $\dfrac14$ in $C([0,1])$.

From a different thread (Regarding integral operators being contractions) it was recommended to use the fact that $u^2 - v^2 = (u+v)(u-v)$ but it's getting me nowhere. Been away from functional analysis for a while and having some trouble getting back into it. Any help is appreciated.


$|\int (u-v) (u+v)| \leq \sqrt {\int (u-v)^{2}} \sqrt {\int (u+v)^{2}}$. For $u$ and $v$ in the given ball $|u+v| \leq \frac 1 2$ so $\sqrt {\int (u+v)^{2}} \leq \frac 1 2$. Of course $\sqrt {\int (u-v)^{2}} \leq \|u-v\|$.

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  • $\begingroup$ @mathhelpmepls There was a typo. I have corrected the answer. $\endgroup$ – Kavi Rama Murthy Feb 5 at 0:34

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