# Show that $\arg\min\limits_{||s||\leq t}\langle p,s\rangle = -t\arg\max\limits_{||s||\leq 1}\langle p,s\rangle$

Consider a real Hilbert space $\mathcal{H}$ with inner product $\langle\cdot,\cdot\rangle$ and corresponding norm $||\cdot||$ induced by the inner product. I cannot see why the following is true (here $p\in\mathcal{H}\backslash\{0\}$),

$\arg\min\limits_{||s||\leq t}\langle p,s\rangle = -t\arg\max\limits_{||s||\leq 1}\langle p,s\rangle$.

We can explicitly calculate the $\min$ and the $\max$. Let us start from the $\min$. We have that for all $s$ such that $\|s\|\le t$ it holds
$$\langle p,s \rangle\ge -|\langle p,s \rangle|\ge -\|p\|\|s\|\ge-t\|p\|,$$
with equality if and only if $s$ and $p$ are linearly dependent, that is $s=\alpha p$ for some $\alpha\in\mathbb{R}$. Choosing $s=-t\frac{p}{\|p\|}$ we get $\langle p,s \rangle=-t\|p\|$, so this $s$ is the only minimizer in $\arg\min_{\|s\|\le t}\langle p,s \rangle$.
Now let us see the $\max$ part. We have that for all $s$ such that $\|s\|\le 1$ it holds
$$\langle p,s \rangle\le|\langle p,s \rangle|\le\|p\|\|s\|=\|p\|,$$
with equality if and only if $s$ and $p$ are linearly dependent, that is $s=\alpha p$ for some $\alpha\in\mathbb{R}$. Choosing $s=\frac{p}{\|p\|}$ we get $\langle p,s \rangle=\|p\|$, so this $s$ is the only maximizer in $\arg\max_{\|s\|\le 1}\langle p,s \rangle$. And comparing the minimizer with the maximizer we get the claim.