Give a sequence of functions $\varphi_n\in \mathcal{C}^\infty(-1,1)$, Cauchy with respect to the Sobolev space $H^1_0$ norm
$$\| \varphi\|_1=\sqrt{\int_{-1}^1 (\varphi')^2+\int_{-1}^1 \varphi^2}$$
which does not converge to a function in $\mathcal{C}^1\left[-1,1\right]$.
I was told an useful approach would be taking a wedge shaped function $f$ in $\mathcal{C^0}(-1,1)$ (i.e. continuous but not differentiable) and considering the convolution $$\psi_n \ast f$$ where $\psi_n$ is a regularizing sequence of functions. It then follows that $$\|(\psi_n\ast f) -f\|_2$$
I am stumped as to what follows, or what I should do next. The weak derivatives do not have the regularity conditions required to apply the above argument to them.
I am new to this subject; any help would be appreciated.
It is true that $\|\psi_{n}\star f-f\|_{L^2}\rightarrow 0$ and $n\rightarrow \infty$ for any $f \in L^2$. If you start with the wedge $f$ as described, then $\|\psi_{n}\star f - f\|\rightarrow 0$ and $$ \|(\psi_{n}\star f)'-f'\|=\|\psi_{n}'\star f-f'\|=\|\psi_{n}\star f'-f'\|\rightarrow 0. $$ This is because, even though $f$ has only a piecewise continuous derivative, \begin{align} \frac{d}{dx}\int \psi_{n}(x-y)f(y)dy & =\int \psi_{n}'(x-y)f(y)dy \\ & = -\int \frac{d}{dy}\psi_{n}(x-y)f(y)dy \\ & = \int \psi_{n}(x-y)f'(y)dy. \end{align} Therefore $\psi_{n}\star f \rightarrow f$ in $H^1$.