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Let $X$ be vector space of real-valued function with continuous derivative on $[0,1]$. Define an inner product on $X$ by $$\langle x, y \rangle = x(0)y(0) + \int_0^1 x^\prime (t) y^\prime (t) dt.$$ Prove that $(X, \langle.,.\rangle)$ is not Hilbert.

This is the last part of a problem in my final exam. In previous parts, we denote $||.||$ be the norm induced by $\langle.,.\rangle$, and consider the norm $||.||_1$ defined by $$||x||_1 = |x(0)|+ \sup_{t\in [0,1]} |x^\prime(t)|.$$ We proved that $||.||_1$ is stronger than $||.||$ (which means if $(x_n)$ converges in $(X,||.||_1)$, then it also converges in $(X, ||.||)$), and then we consider $(x_n)$ given by $$x_n(t) = n\left(\frac{t^{n+1}}{n+1} - \frac{t^{n+2}}{n+2}\right)$$ which converges to $0$ in $(X, ||.||)$ but does not converge to $0$ in $(X, ||.||_1)$.

Thank you very much.

Tien Kha Pham
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    The map $f\mapsto (f(0),f')$ is a unitary isomorphism $X\to \mathbb R\oplus C[0,1]$ where $C[0,1]$ has the scalar product $\int f g, dx$. You can read here why $C[0,1]$ is not complete, so your space cannot be a Hilbert space. – s.harp Dec 31 '16 at 11:00
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    Choose a sequence ${ y_n }$ of piecewise linear continuous functions $0 \le y_n \le 1$ that converges in $L^2$ to $\chi_{[\frac{1}{2},1]}$. Then define $x_n(t)=\int_{0}^{t}y_n(u)du$. ${x_n}$ converges uniformly to $\int_{0}^{t}\chi_{[\frac{1}{2},1]}(u)du$, and each $x_n$ is continuously diffierentiable with $x_n'=y_n$. – Disintegrating By Parts Dec 31 '16 at 16:49

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