The parallelogram identity

Keywords: pre-Hilbert space · supremum norm · parallelogram identity · counterexample

Exercise 1 : The parallelogram identity

Question 1

Let

E = \mathcal{C}^0([0,1],\mathbb{C})

be the space of continuous complex-valued functions on

[0,1]

, equipped with the norm

\|f\|_\infty = \sup_{t \in [0,1]} |f(t)|.

Show that $(E, \|\cdot\|_\infty)$ is not a pre-Hilbert space.

Hint

You must show that the norm cannot be induced by an inner product, and hence find a counterexample to the parallelogram identity. Try using simple functions.

Solution

It suffices to show that

\|\cdot\|_\infty

violates the parallelogram identity:

\|f + g\|^2 + \|f - g\|^2 = 2\bigl(\|f\|^2 + \|g\|^2\bigr).

Set $f(t) = 1$ and $g(t) = t$ . We compute:

\|f\|_\infty = 1, \quad \|g\|_\infty = 1, \quad \|f+g\|_\infty = \sup_{t\in[0,1]}|1+t| = 2, \quad \|f-g\|_\infty = \sup_{t\in[0,1]}|1-t| = 1.

It follows that:

\|f+g\|_\infty^2 + \|f-g\|_\infty^2 = 4 + 1 = 5 \neq 4 = 2\bigl(\|f\|_\infty^2 + \|g\|_\infty^2\bigr).

The parallelogram identity is therefore not satisfied, which implies that $\|\cdot\|_\infty$ cannot be induced by any inner product. The space $(E,\|\cdot\|_\infty)$ is not a pre-Hilbert space.