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Hi I'm stuck on the following problem of Haim's functional analysis book.

Let $C\subseteq H$ ($H$ a Hilbert space) be a non-empty closed convex subset and let $T:C\rightarrow C$ be a non linear contraction, i.e., $|Tu-Tv|\leq|u-v|$ $\forall u,v\in C$

  1. Let $(u_{n})$ be a sequence in $C$ such that $u_{n}\rightharpoonup u$ weakly and $u_{n}-Tu_{n}\rightarrow f$ strongly. Prove that $u-Tu=f$. (HINT: start with the case $C=H$ and use the inequality $<(u-Tu)-(v-Tv),u-v>\geq0$)

  2. Deduce that if $C$ is bounded then $T$ has a fixed point. (HINT: Consider $T_{\varepsilon}u=(1-\varepsilon)Tu+\varepsilon a$ with $a\in C$ fixed and $\varepsilon>0$.)

Thanks for your help.

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1 Answer 1

Here's a sketch of a solution.

  1. As in the hint, first suppose $C = H$. Since $T$ is a contraction, we get the inequality $((u-Tu) - (v-Tv), u-v) \geq 0$ for all $u, v \in H$. Apply this inequality with $u_n$ and $u + tv$, where $v \in H$, $t \in \mathbb{R}$. We get \begin{equation} ((u_n - Tu_n) - (u+tv - T(u+tv)), u_n - u - tv) \geq 0. \end{equation} Since $(u_n - Tu_n)$ converges strongly to $f$ and $u_n - u$ converges weakly to $0$ as $n \to \infty$, we get \begin{equation} (f-(u+tv - T(u+tv)), -tv) \geq 0. \end{equation} Now observe that since $\lvert T(u+tv) - Tu \rvert \leq \lvert tv \rvert \to 0$ as $t \to 0$, we have $T(u+tv) \to Tu$ as $t \to 0$.
    Dividing the inequality above by $-t$ and sending $t \to 0^-$ and $t \to 0^+$, we obtain \begin{equation} (f-(u-Tu),v) = 0, \, \forall v \in H. \end{equation} Hence $f = u -Tu$. For the general case where $C \neq H$, apply the above result to $\tilde{T} = T \circ P_C: H \to H$, where $P_C$ is the projection onto $C$. (Check that $\tilde{T}$ is a contraction!)

  2. It is easy to show that $T_\epsilon$ is a strict contraction for all $\epsilon > 0$, so the contracting mapping principle gives a fixed point $u_\epsilon \in C$ of $T_\epsilon$. If $C$ is bounded, then $(u_\epsilon)$ has a weakly convergent subsequence $u_\epsilon \rightharpoonup u$, since $C$ is reflexive. Moreover, one can check that $(u_\epsilon - Tu_\epsilon)$ converges strongly to $0$ as $\epsilon \to 0$. So by part (1), $u - Tu = 0$ and $u$ is a fixed point of $T$.

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