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Let us consider a Hilbert space $\mathbb{R}^{n}$ equipped with a dot product $x\cdot y = \sum_{i=1}^{n}x_{i}y_{i}$. Next, let $S$ be a convex closed subset of $\mathbb{R}^{n}$ and let $s$ be a projection of $x$ onto $S$.

First, from the property of orthogonal projection, it follows that $$ \sum_{i=1}^{n}x_{i}s_{i} = \sum_{i=1}^{n}s_{i}s_{i} $$

Next, let $y$ be some vector from $S$. What is the relation between $\sum_{i=1}^{n}x_{i}y_{i}$ and $\sum_{i=1}^{n}s_{i}y_{i}$?

Is it always $$ \sum_{i=1}^{n}x_{i}y_{i} \leq \sum_{i=1}^{n}s_{i}y_{i} $$ ?

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The difference in the two terms you are trying to compare is $$ (x - s) \cdot y = \sum_{i=1}^n (x_i - s_i) y_i $$ If $x = s$ (that is, if $x \in S$), then it is always zero and the two terms are always equal. If they are different, this can have any value by varying $y$. It can be positive by setting $y = x - s$ or it can be negative by setting $y = -(x - s)$.

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  • $\begingroup$ wait... would $y=x-s$ belong to S? $\endgroup$
    – LrM
    Mar 29, 2021 at 17:46
  • $\begingroup$ Missed the fact that $y \in S$ but the point still stands. If $S$ is a ball centered at the origin, then some multiple of $x-s$ and $-(x-s)$ will be in $S$. $\endgroup$
    – MBW
    Mar 30, 2021 at 14:37
  • $\begingroup$ @LrM I also just realized that there is something you can say about $(x - s) \cdot (y - s)$ which is that $(x - s) \cdot (y - s) \leq 0$ $\endgroup$
    – MBW
    Mar 30, 2021 at 15:57
  • $\begingroup$ well, the inequality in the last message is a property of projection onto any convex closed set. This is not what is in the question. $\endgroup$
    – LrM
    Mar 31, 2021 at 0:24

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