Problem: On $C^1[0,1]$, we define an inner product by

$$\langle x, y \rangle = x(0)y(0) + \int_0^1 x^\prime (t) y^\prime (t) dt.$$ Prove that $(C^1[0,1],\langle \cdot,\cdot \rangle)$ is not Hilbert space.

My attempt: Suppose that $(C^1[0,1],\langle \cdot,\cdot \rangle)$ is Hilbert space.

Next, we denote $\Vert \cdot \Vert$ be the norm induced by $\langle \cdot,\cdot \rangle$. Thus, for all $x\in C^1[0,1]$ we have $$\Vert x \Vert= \sqrt{x^2(0)+\int_{0}^{1}[x'(t)]^2dt}.$$

Consider $(x_n)_n \subset C^1[0,1]$ defined by $x_n(t)=\dfrac{t^{n+1}}{n+1}$. I will prove that $(x_n)_n$ is Cauchy sequence as following $$\lim_{n\to+\infty}\Vert x_{n+p}-x_n \Vert^2= \lim_{n\to+\infty}\int_{0}^{1}t^{2n}(t^p-1)^2dt=\int_{0}^{1}\lim_{n\to+\infty}t^{2n}(t^p-1)^2dt=0.$$ Since, $(C^1[0,1],\langle\cdot,\cdot\rangle)$ is Hilbert space, we have $(x_n)$ converges to $x_0 \in (C^1[0,1],\Vert\cdot\Vert)$. Now, I just have to show that $x_0 \notin C^1[0,1]$ then the problem will be solved but I have stucked.

Thanks for any help.

P/S: In addition, I also try to solve this problem by choosing a sequence that is the Cauchy sequence but not convergent in $(C^1[0,1],\Vert \cdot \Vert)$ but it does not work.

  • $\begingroup$ Your sequence $(x_n)_n$ simply converges to $0$ in $(C^1[0,1],\|\cdot\|)$. $\endgroup$ Commented Dec 18, 2020 at 10:41

1 Answer 1


Here is a different solution. Your norm is $$\|f\| = \sqrt{|f(0)|^2+\|f'\|_2^2}$$ and we know that $(C[0,1],\|\cdot\|_2)$ is not complete with the non-convergent Cauchy sequence given with $$f_n(x) := \begin{cases} 0, & \text{if $x \in \left[0,\frac1{2}\right]$}\\ n\left(x-\frac{1}{2}\right), & \text{if $x \in \left[\frac1{2},\frac1{2}+\frac1n\right]$}\\ 1, & \text{if $x \in \left[\frac1{2}+\frac1n,1\right]$} \end{cases}$$ Therefore it makes sense to define $$g_n(x):=\int_0^x f_n(t)\,dt$$ which is a $C^1[0,1]$-function with $g_n' = f_n$. For $m \ge n$ we have \begin{align} \|g_m-g_n\|^2 &= |g_m(0)-g_n(0)|^2 + \int_0^1 |g_m'(t)-g_n'(t)|^2\,dt\\ &= \int_0^1 |f_m(t)-f_n(t)|^2\,dt\\ &= \int_{\frac12}^{\frac12+\frac1m} (m-n)^2\left(t-\frac12\right)^2\,dt + \int_{\frac12+\frac1m}^{\frac12+\frac1n}\left(1-n\left(x-\frac12\right)\right)^2\,dt\\ &= (m-n)^2\int_{0}^{\frac1m} t^2\,dt + \int_{\frac1m}^{\frac1n}(1-nt)^2\,dt\\ &= \frac{(m-n)^2}{3m^3}+\frac{\left(1-\frac{n}{m}\right)^3}{3n}\\ &\le \frac1{3m}+\frac1{3n} \xrightarrow{m,n\to\infty} 0 \end{align}

so $(g_n)_n$ is a Cauchy sequence in $(C^1[0,1],\|\cdot\|)$. Suppose that $g_n \xrightarrow{\|\cdot\|} g$ for some $g \in C^1[0,1]$.

In particular this means $g_n(0) \to g(0)$ and $g_n' \xrightarrow{L^2} g'$ which implies that $g'$ is the $L^2$-limit of $(f_n)_n$. We have $$\left\|f_n-\chi_{\left\langle\frac12,1\right]}\right\|_2^2 = \int_{\frac12}^{\frac12+\frac1n} \left(1-n\left(x-\frac12\right)\right)^2\,dt = \int_0^{\frac1n} (1-nx)^2\,dt = \frac1{3n} \xrightarrow{n\to\infty} 0$$ so $g' = \chi_{\left\langle\frac12,1\right]}$ almost everywhere. Since $g(0)=0$, it follows that $$g(x) = \int_0^x g'(t)\,dt = \left(x-\frac12\right)\chi_{\left\langle \frac12,1\right]}(x)$$ which is not differentiable at $x=\frac12$. This is a contradiction.


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