Most answers to this question use the paralellogram law but I am wondering if the following holds.

Show that the maximum norm on $C[a,b]$ is not induced by an inner product.

Assume the maximum norm is induced by an inner product. Consider the function $f(x)=x^2-1$ on $[-1,1]$. $\|f\|^2_\text{max}=\langle f, f \rangle =0$ but $f\neq 0$ which contradicts bi-linearity.

Now consider the sequences $f=(1,-1,1,0,0\dots)$ and $g=(-1,1,-1,0,0\dots)$

$\|f+g\|^2+\|f-g\|^2=0\neq 2*3^2+2*3^2=2\|f\|^2+2\|g\|^2$

which implies the parallelogram law does not hold.

  • 2
    $\begingroup$ The maximum norm is the maximum of $|f(x)|$ on the interval, not the maximum of $f(x)$. So for $f(x)=x^2-1$ on $[-1, 1]$ the maximum norm is $1$. $\endgroup$
    – angryavian
    May 14 at 3:02

1 Answer 1


Consider $C[-1,1].$ Assume $$\|f\|^2_\max=\langle f,f\rangle.$$ Then for $f_a(x)=1-ax^2,$ $0< a\le 1,$ we get $$\displaylines{1=\langle 1-ax^2,1-ax^2\rangle \\ =\langle 1,1\rangle -2 \Re\langle 1,x^2\rangle \,a+\langle x^2,x^2\rangle\,a^2=:\varphi(a)}$$ The quadratic function $\varphi(a)$ is constant for $0< a\le 1.$ Thus $\langle x^2,x^2\rangle =0,$ which gives a contradiction.


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