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Let $a\in\mathbb{R}^{d}$ and $K$ be an arbitrary subset of $\mathbb{R}^{d}$. My question is related to the following optimization problem: \begin{equation} \max_{x\in\mathbb{R}^{d}}~a^{\top}x\quad \text{s.t.}~~x\in K \end{equation} Is it true that the solution to the above problem is just the projection of $a$ onto $K$? Does this at least hold under some additional restriction on $K$, like convexity? If yes, how can this be shown?

Related, but unanswered: Maximizing an inner-product over a convex set.

nemo
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1 Answers1

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No, it’s not just the projection, even when $K$ is convex. Let $K$ be the unit circle centered at the origin, and let $a=(2,2)$. Then the optimal solution to your proposed optimization problem is $x^*=(-1,-1)$, while the projection of $a$ onto $K$ is $(1,1)$.

Looks like the post got edited so here’s a different example. Let $K$ be the unit square in $\mathbb{R}^2$, and let $a=(1/2,1/2)$. Then the maximizer is $x^*=(1,1)$, while the projection of $a$ onto $K$ is the vector $a$ itself.

For an example with $a\not\in{K}$, let $K$ again be the unit square, and take $a=(-\varepsilon,\varepsilon)$ for some $0<\varepsilon<1$. Then the maximizer is $x^*=(0,1)$, while the projection of $a$ onto $K$ is $(0,\varepsilon)$

One more example, where $x^*$ is not a scalar multiple of $a$: let $K$ be the circle in $\mathbb{R}^2$ with radius $1$ centered at $(1,1)$ and take $a=(0,1)$. Then the maximizer is $x^*=(1,2)$.

David M.
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  • How can $(1,1)$ be the maximizer? I'm assuming by unit square you mean the $\ell_{1}$ ball, $(1,1)$ doesn't lie on it. – nemo Mar 13 '19 at 08:42
  • @nemo Sorry should have been clearer—by unit square I mean the square with corners at $(0,0)$, $(0,1)$, $(1,0)$ and $(1,1)$. – David M. Mar 13 '19 at 08:43
  • Okay, thanks for clarifying. – nemo Mar 13 '19 at 08:51
  • One more question: is the $\textit{any}$ relation between the maximizer and projection of $a$ onto $K$, say wrt Euclidean distance? Also, seems like $K$ must be bounded in some way, otherwise the inner product can be made arbitrarily large, as in your last example – nemo Mar 13 '19 at 13:20
  • @nemo If $K$ is unbounded, then the optimization problem may or may not have an infinite solution. – David M. Mar 13 '19 at 14:22
  • @nemo And there may be some kind of relationship (especially if $K$ exhibits some kind of symmetry) but it’s not clear to me what a general result would look like. Someone else on here may know – David M. Mar 13 '19 at 14:32