Show that the sup-norm is not derived from an inner product

Question

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 $\lVert{\cdot} \rVert _{\infty}$

Answer 1

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.

Answer 2

$\|u+v\|^2+\|u−v\|^2=2\|u\|^2+2\|v\|^2$

the parallelogram identity the parallelogram identity is necessary and sufficient for a given norm to be derived from an inner product.

You can show that in your case it does not hold.

Answer 3

If the norm came from a scalar product, then the parallelogram law would hold. But it is easy to find two functions $f$ and $g$ such that $$2||f||_\infty^2 + 2||g||_\infty^2 \neq ||f+g||_\infty^2 + ||f-g||_\infty^2.$$

Answer 4

Explicit continuous functions that work on the unit interval (i.e. demonstrate the failure of the parallelogram law holding) are $f(x) = 1-x$ and $g(x) = x$.

$2||f||^2_\infty = 2$ and $2||g||^2_\infty = 2$, while $||f+g||^2_\infty = 1$ and $||f-g||^2_\infty = 1$.