Given $\vec y \in \mathbb{R}^n$, compute the operator norm $\lVert f\rVert_{op}$, where $f(\vec x)=\langle\vec x, \vec y\rangle=\sum_{k=1}^n x_ky_k$.
My attempt: we know that $\lVert f\rVert_{op}=\sup\{\lVert f(\vec x)\rVert_{2}: x\in S^{n-1}\}$, where $S^{n-1}=\{\vec x\in\mathbb{R}^n: \lVert \vec x\rVert_2=1\}$. Note $\lVert f(\vec x)\rVert_2=\sqrt{\langle\vec x, \vec y\rangle^2}=\langle\vec x, \vec y\rangle$ and $\lVert \vec x\rVert_{2}=\sqrt{\langle\vec x, \vec x\rangle^2}=\langle\vec x, \vec x\rangle$, thus $\lVert f\rVert_{op}=\sup\{\langle\vec x, \vec y\rangle:\; \langle\vec x, \vec x\rangle=1\}$. By the Cauchy-Schwarz inequality, $\langle\vec x, \vec, y\rangle^2\leq \langle\vec x, \vec x\rangle\langle\vec y, \vec y\rangle$, hence $\lVert f(\vec x)\rVert_2 \leq \sqrt{\langle\vec x, \vec x\rangle\langle\vec y, \vec y\rangle}=\sqrt{\langle\vec y, \vec y\rangle}$.
My question: so we know that $\lVert f(\vec x)\rVert_2$ is bounded above by $\sqrt{\langle\vec y, \vec y\rangle}$, but that does not necessarily mean that $\sqrt{\langle\vec y, \vec y\rangle}$ is the supremum. How exactly do I find the supremum of $\{\langle\vec x, \vec y\rangle:\; \langle\vec x, \vec x\rangle=1\}$?