Let $V = \mathcal{C}[0,1]$ with inner product $\langle v,w\rangle =\int_{0}^1 v(x)w(x) \mathrm{d}x$. Find a functional $f\in V^*$ for which there does not exist a vector $v\in V$ satisfying $f(w) = \langle v,w\rangle$.
I tried to use $f(w) = w(\frac{1}{2})$, but I cannot find a contradiction. Any help?
Try something like $$ f(w)=\int_0^{a}w(x)\,dx\text{ where }0<a<1. $$ $$ |f(w)|=\Bigl|\int_0^aw(x)\,dx\Bigr|\le\Bigl(\int_0^a|w(x)|^2\,dx\Bigr)^{1/2}\Bigl(\int_0^adx\Bigr)^{1/2}\le a^{1/2}\,\|w\|, $$ so that $f$ is continuous on $V$ with the norm coming from the inner product ($\|w\|^2=\langle w,w\rangle$.) Let $v(x)=1$ if $0\le x\le a$, $v(x)=0$ if $a<x\le1$. Then $f(w)=\langle w,v\rangle$, but $v\not\in V$.