Show that the space $C[0, 1]$ of real-valued continuous functions on the unit interval $[0, 1]$ with the sup norm $$ \|f\|=\sup\{|f(x)|:\ x\in[0,1]\} $$ is not a Hilbert space with respect to any inner product .
My attempts: as I have to find a Cauchy sequence $(f_n)_n$ which converges to a function $f$ which is not continuous, but I can't construct such a sequence $(f_n)_n$.
Any norm coming from an inner product satisfies the parallelogram identity. If for instance you take $f(x)=x$, $g(x)=1$, then $$ \|f+g\|^2+\|f-g\|^2=5, $$ while $$ 2\|f\|^2+2\|g\|^2=4. $$
It suffices to show that the parallelogram law is not satisfied. Take $f(x)=x$, $x\in[0,1]$, and $g(x)=1$, $x\in[0,1]$. Then $2(\|f\|_\infty^2+\|g\|_\infty^2)=4$, but $\|f+g\|_\infty^2+\|f-g\|_\infty^2=5$.
There is a theorem that says that a norm $(X,\|.\|)$ comes from an inner product if and only if the parallellogram law is satisfied for all elements in $X$: $$ 2\|x\|^2+2\|y\|^2 = \|x+y\|^2+\|x-y\|^2.$$
