# Simple identity for inf and sup of inner product

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 $$S$$.