Let $f$ be an element in the Sobolev space $H^1(\mathbb{R})=W^{1,2}(\mathbb{R})$. For every $h>0$, let $$f_h(x)=\frac{1}{h}(f(x+h)-f(x))$$ Prove that $$\lim_{h\to 0^+}\left\Vert f'-f_h\right\Vert_{L^2(\mathbb{R})}=0$$
My proof seems to be unnecessarily long and anyone who can kindly check if there are any mistakes will be appreciated!
Let $\mathcal{D}$ denote the set of all smooth, compactly supported functions $\varphi:\mathbb{R}\to \mathbb{R}$. It is well-known that $\mathcal{D}$ is dense in $H^1(\mathbb{R})$. So there is a sequence $(\varphi_n)$ in $\mathcal{D}$ such that $$\lim_{n\to \infty}\left\Vert \varphi_n-f\right\Vert_{H^1(\mathbb{R})}=\lim_{n\to \infty}\left(\left\Vert \varphi_n-f\right\Vert_{L^2(\mathbb{R})}+\left\Vert \varphi_n'-f'\right\Vert_{L^2(\mathbb{R})}\right)=0$$ It follows that $$\lim_{n\to \infty}\left\Vert \varphi_n-f\right\Vert_{L^2(\mathbb{R})}=\lim_{n\to \infty}\left\Vert \varphi_n'-f'\right\Vert_{L^2(\mathbb{R})}=0$$ So both $\varphi_n$ and $(\varphi_n')$ are convergent, hence Cauchy, in $L^2(\mathbb{R})$. Hence, for any $\epsilon>0$, there exists an $N_1\in \mathbb{N}$ such that $$i,j>N_1\Rightarrow\left\Vert \varphi_i'-\varphi_j'\right\Vert_{L^2(\mathbb{R})}<\epsilon$$ For every $h>0$ and every $n\in \mathbb{N}$, define the function $$\varphi_{h,n}(x)=\frac{1}{h}(\varphi_n(x+h)-\varphi_n(x))$$ Now let $n$ be fixed. By the mean value theorem, for every $x$, there exists some $y\in (x,x+h)$ such that $$\left|\varphi_n'(x)-\varphi_{h,n}(x)\right|^2=\left|\varphi_n'(x)-\varphi_n'(y)\right|^2\leq 2\left\Vert \varphi_n\right\Vert_{C^1(\mathbb{R})}^2=M_n$$ So the sequence $(\left|\varphi_n'-\varphi_{h,n}\right|^2)_h$ is dominated by the Lebesgue integrable function $g_n=M_n\cdot\chi_{\text{support}(\varphi_n)}$ (which is a scalar multiple of a characteristic function). By the dominated convergence theorem, $$ \begin{aligned} \lim_{h\to 0}\int_{\mathbb{R}}\left|\varphi_n'(x)-\varphi_{h,n}(x)\right|^2dx &=\int_{\mathbb{R}}\lim_{h\to 0}\left|\varphi_n'(x)-\varphi_{h,n}(x)\right|^2dx \\ &=\int_{\mathbb{R}}\left[\lim_{h\to 0}\left|\varphi_n'(x)-\varphi_{h,n}(x)\right|\right]^2dx \\ &=0 \end{aligned} $$ So for every $n\in \mathbb{N}$ we have $$\lim_{h\to 0}\left\Vert \varphi_n'-\varphi_{h,n}\right\Vert_{L^2(\mathbb{R})}=0$$ Furthermore, when $h$ is fixed, $(\varphi_{h,n})_n$ converges to $f_h$ in $L^2(\mathbb{R})$ because $\varphi_n$ converges to $f$ in $L^2(\mathbb{R})$. In particular, $(\varphi_{h,n})_n$ is Cauchy in $L^2(\mathbb{R})$. Hence, for any $\epsilon>0$, there exists an $N_2\in \mathbb{N}$ such that $$i,j>N_2\Rightarrow\left\Vert \varphi_{h,i}-\varphi_{h,j}\right\Vert_{L^2(\mathbb{R})}<\epsilon$$ Finally, we know that for arbitrary $m,n,p\in \mathbb{N}$, $$ \left\Vert f'-f_h\right\Vert_{L^2(\mathbb{R})} \leq \left\Vert f'-\varphi_m'\right\Vert_{L^2(\mathbb{R})}+ \left\Vert \varphi_m'-\varphi_n'\right\Vert_{L^2(\mathbb{R})}+ \left\Vert \varphi_n'-\varphi_{h,n}\right\Vert_{L^2(\mathbb{R})}+ \left\Vert \varphi_{h,n}-\varphi_{h,p}\right\Vert_{L^2(\mathbb{R})}+ \left\Vert \varphi_{h,p}-f_h\right\Vert_{L^2(\mathbb{R})} $$ Fix an $\epsilon>0$.
(i) Choose an $m_0>N_1$ and an $n_0>\max\{N_1,N_2\}$ such that $$\left\Vert f'-\varphi_{m_0}'\right\Vert_{L^2(\mathbb{R})}<\frac{\epsilon}{5};\qquad \left\Vert \varphi_{m_0}'-\varphi_{n_0}'\right\Vert_{L^2(\mathbb{R})}<\frac{\epsilon}{5}$$
(ii) There exists an $h_0>0$ such that whenever $0<h<h_0$, $$\left\Vert \varphi_{n_0}'-\varphi_{h,n_0}\right\Vert_{L^2(\mathbb{R})}<\frac{\epsilon}{5}$$
(iii) For any fixed $h\in (0,h_0)$, choose a $p_0>N_2$ such that $$\left\Vert \varphi_{h,n_0}-\varphi_{h,p_0}\right\Vert_{L^2(\mathbb{R})}<\frac{\epsilon}{5};\qquad \left\Vert \varphi_{h,p_0}-f_h\right\Vert_{L^2(\mathbb{R})}<\frac{\epsilon}{5}$$ We conclude that for every $h\in (0,h_0)$, $$\left\Vert f'-f_h\right\Vert_{L^2(\mathbb{R})}<\epsilon$$ giving the result.