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I have a question about the following result from a paper that I am currently reading.


Let $X$ be a Hilbert space with inner product $\langle \cdot, \cdot\rangle$, $C$ be a finite dimensional subspace of $X$, and $S$ be a bounded subset of $X$. Suppose a nonlinear mapping $F:X\to X^*$ satisfies: $$\langle F(x)-F(y),z \rangle\le c|x-y||z|,\quad x,y,z\in S,$$ then the author claims that $$\langle PF(x)-PF(y),z \rangle\le c|x-y||z|\quad x,y,z\in S,$$ where $P:X\to C$ is the projection operator onto $C$.


I was wondering whether this is correct. I thought this follows from $$\langle Px,z \rangle \le \langle x,z \rangle ,x,z\in S$$ which is incorrect, since in that paper, there is no inclusion relation between $S$ and $C$.

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  • $\begingroup$ How does $PF(x)$ make sense? If $F:X \to X^*$ and $P:X \to C$ then $F(x) \in X^*$, and so what does $P$ do to an element of $X^*$? (Assuming that $PF(x) = P(F(x))$) $\endgroup$ Aug 13, 2017 at 17:50
  • $\begingroup$ I think that it is by identifying $X^*$ as $X$ $\endgroup$
    – John D
    Aug 13, 2017 at 17:52
  • $\begingroup$ @FlybyNight I think the author identified $X$ as $X^*$. $\endgroup$
    – John
    Aug 13, 2017 at 17:55

1 Answer 1

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Consider the dual operator of P.

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  • $\begingroup$ I see. And use the fact $\|P^*\|=\|P\|$. $\endgroup$
    – John
    Aug 13, 2017 at 18:00

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