Consider the $\mathbb R-$linear space $B(X,\mathbb C)$ of all bounded functions from $X\to\mathbb C$ for some $X\ne\emptyset.$ Then $\displaystyle||.||:B(X,\mathbb C)\to\mathbb R:f\mapsto\sup_{x\in X}|f(x)|$ is a norm on $B(X,\mathbb C).$
Is't possible to define a real inner product $\langle,\rangle: B(X,\mathbb C)\times B(X,\mathbb C)\to\mathbb R$ such that it generates the above norm?
If $X$ consists of more than one point then it is not going to happen. Assume $a,b\in X$. Then you can have $f(x)=0$, for $x\neq b$, $g(x)=0$, for $x\neq a$, and $f(b)=g(a)=||f||=||g||=1$. Then
$$||f+g||^2+||f-g||^2=|f(a)+g(a)|^2+|f(a)-g(a)|^2=2||f||^2=2\neq4=2||f||^2+2||g||^2.$$
For $X=\{a\}$ then we have
\begin{align}||f+g||^2+||f-g||^2&=|f(a)+g(a)|^2+|f(a)-g(a)|^2\\&=(f(a)+g(a))(\overline{f(a)}+\overline{g(a)})+(f(a)-g(a))(\overline{f(a)}-\overline{g(a)})\\&=2f(a)\overline{f(a)}+2g(a)\overline{g(a)}\\&=2|f(a)|^2+2|g(a)|^2\\&=2||f||^2+2||g||^2.\end{align}
Therefore it comes from an scalar product.
More directly, it comes from the scalar product $$\left<f,g\right>:=f(a)\overline{g(a)}.$$