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Let $H$ be a hilbert space which is the completion of the set:

$$C^1[0,1] = \{ f\in C[0,1]: \exists f'(t)\in C[0,1]\}$$

in the norm $\|f\|^2 = \int_0^1 |f'(t)|^2 dt+ \int_0^1 |f(t)|^2 dt$

Then I want to prove that for any $f\in H$, we can represent $f$ by a continuous function.

I.e. $\{f_n(x)\}_{n=1}^\infty$ which is a Cauchy sequence in $H$, converges uniformly to some $f\in C[0,1]$


So for $\epsilon>0$, $n>m>N$ $\|f_m-f_n\|<\epsilon$

I worked backwards here, but now I am stuck: $$\left|\int_0^1 f_m(t)-f_n(t) dt \right|\leq\int_0^1 |f_m(t)-f_n(t)|dt\leq\sqrt{\int_0^1 |f_m(t)-f_n(t)|^2 dt}\leq\sqrt{\int_0^1 |f'_m(t)-f_n'(t)|^2 dt + \int_0^1 |f_m(t)-f_n(t)|^2 dt}<\epsilon$$

How do I progress?

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Define \begin{align} \|h\|_{L^2}&=\left(\int_{0}^{1}|h|^2dt\right)^{1/2},\;\;\; h\in L^2[0,1]\\ \|f\|_{H} &= \left(\|f\|_{L^2}^2+\|f'\|_{L^2}^2\right)^{1/2},\;\;\; f\in C^1[0,1]. \end{align} Then there exists a constant $C$ such that, for all $f\in C^1[0,1]$ and $x\in [0,1]$, \begin{align} f(x)(1-x)&=\int_{x}^{1}f(t)(-1)dt+\int_{x}^{1}f'(t)(1-t)dt \\ |f(x)|(1-x)&\le \|f\|_{L^2}\|1\|_{L^2}+\|f'\|_{L^2}\|1-t\|_{L^2} \\ &\le \left(\|f\|_{L^2}^2+\|f'\|_{L^2}^2\right)^{1/2}\left(\|1\|_{L^2}^2+\|1-x\|_{L^2}^2\right)^{1/2} \\ & = C\|f\|_{H} \end{align} Similarly, there exists a constant $D$ such that, for all $f\in C^1[0,1]$ and $x\in[0,1]$, $$ |f(x)|x \le D\|f\|_{H} $$ Adding these gives $$ |f(x)| \le (C+D)\|f\|_{H} $$

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