If $X$ has more than one element, then the norm $\|f\|=\sup_{x\in X}|f(x)|$ doesn't come from a inner product in $B(X;\mathbb{R}) := \{f \mathpunct{:} X\to \mathbb{R}$ such that $f$ is bounded$\}$
That is, there is no inner product $\langle-,-\rangle$ such that $\langle f,f\rangle = \|f\|^2$.
I thought of doing this in two ways: Set $\langle f,f \rangle = \|f\|^2$ and then getting a contradiction to the definition of inner product. Or showing a counterexample of the parallelogram law. In both cases I'm stuck on how to use that $X$ has more than one element. Any hint or idea will be appreciated.
If $X$ contains elements $a\neq b$, define $f(x) = \delta_{xa}$ and $g(x) = \delta_{xa} + \delta_{ba}$. Then $$ 2(\|f\|_\infty^2 + \|g\|_\infty^2) = 2 \neq 5 = \|f+g\|_\infty^2 + \|f-g\|_\infty^2 $$ so $\|\cdot\|_\infty$ doesn't satisfy the parallelogram law and hence isn't generated by an inner product.