Let $V$ be an inner product space, $S \subseteq V$, and $y \in V$.

I believe the following identity is true: $$ \newcommand{\inner}[1]{\langle #1 \rangle} \inf_{x \in S} \inner{x,y} = -\sup_{x \in S} \inner{x,-y}. $$

If $S$ satisfies sufficient conditions for the $\arg \min$ to exist (closed, compact?), then it is easy:

Let $x^\star = \arg \min_{x \in S} \inner{x,y}$. Then $$ \min_{x \in S} \inner{x,y} = \inner{x^\star, y} = -\inner{x^\star, -y} \geq -\max_{x \in S} \inner{x, -y}. $$ Let $x_\star = \arg \max_{x \in S} \inner{x,-y}$. Then $$ -\max_{x \in S} \inner{x,-y} = -\inner{x_\star, -y} = \inner{x_\star, y} \geq \min_{x \in S} \inner{x, y}. $$ But this proof technique cannot be applied version with $\inf$ and $\sup$ because it would require something like $\arg \sup$ to exist.

I believe we can assert the existence of $\arg \sup$ by the continuity of the inner product, if we also require completeness of $V$, and maybe some other conditions that I am forgetting.

However, I think the theorem is still true even if the space is not complete...

Is there a better proof technique that avoids this issue?


If $T$ is any set of real numbers and $-T$ denotes $\{-x:x\in T\}$ then $\inf T=-\sup (-T)$. Just take $T=\{\langle x, y \rangle : y \in S\}$.

The only property of inner product you require is $ -\langle x, y \rangle=\langle x, -y \rangle$ and you don't have to know any special property of the set $S$.


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