# Does this show that the maximum norm on $C[a,b]$ is not induced by an inner product?

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.

• 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 $1$. May 14, 2022 at 3:02

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.