Show that the sup-norm is not derived from an inner product I am trying to show that the norm $$\lVert{\cdot} \rVert _{\infty}=\sup_{t \in R}|x(t)|$$  does not come from an inner product (the norm is defined on all bounded and continuous real valued functions).
I tried to show that the inner product does not hold by using the conjugate symmetry, linearity and non-degenerancy conditions. But I am unsure of how to do it for the norm $\lVert{\cdot} \rVert _{\infty}$

To simplify my answer, I'll ignore the "continuous" requirement and assume there is an appropriate inner product for that norm.
Let $b$ be a real number, $$f(x) = \left\{ {\begin{array}{*{20}{c}}
  {1,}&{x = 0} \\ 
  {0,}&{x \ne 0} 
\end{array}} \right.$$
and $$g(x) = \left\{ {\begin{array}{*{20}{c}}
  {1,}&{x = 2} \\ 
  {0,}&{x \ne 2} 
\end{array}} \right.$$
Then we would have $${\left\| {f + bg} \right\|^2} = {\left\| f \right\|^2} + {b^2}{\left\| g \right\|^2} + 2b\left\langle {f,g} \right\rangle $$
which varies quadratically for varying $b$. However, in our case, $\left\| {f + bg} \right\|$ is 1 for $\left| b \right| \leqslant 1$  ($f$ dominates) and is $\left| b \right|$ for $\left| b \right| > 1$ ($bg$ dominates). This is the wrong kind of variation, so the inner product must not exist.
You can easily make continuous functions $f$ and $g$ that behave similarly.
