When reading a paper, I encountered the following statement:
''Given a bounded function $f:\mathbb{R}\rightarrow\mathbb{R}$ with $f''\in L^2(\mathbb{R})$ and $\epsilon>0$ we can consider the central difference method with discretisation distance $\epsilon$ as $$(\Delta_\epsilon f)(x)=\frac{f(x+\epsilon)+f(x-\epsilon)-2f(x)}{\epsilon^2}.$$ Then we have the limit $$\lim\limits_{\epsilon\downarrow 0}||\Delta_\epsilon f-f''||_{L^2(\mathbb{R})}=0\text{.''}$$
As it happens, the author claims this statement follows easily and does not provide a proof. Intuitively the statement makes sense of course. And I can prove the pointwise limit, at least if $f$ is $C^2$. However, I tried several things, but I cannot seem to prove the norm convergence. Can anyone help me here? Thank you in advance.
Edit: after some discussion, I found an error, which I am not able to fix. I have adjusted the proof so that it at least works for the special case that $\lim\limits_{x\rightarrow\pm\infty}f(x)$ and $\lim\limits_{x\rightarrow\pm\infty}f(x)’$ exist.
Original post: After working on it for quite some time, I found the solution.
The Fourier transform of $\Delta_\epsilon f-f''$ is the given by \begin{equation*} \begin{array}{lcl}\int\limits_{-\infty}^\infty\frac{1}{\epsilon^2}\big(e^{2\pi i \epsilon \xi}+e^{-2\pi i \epsilon \xi}-2\big)e^{-2\pi i \xi x}f(x)-e^{-2\pi i \xi x}f''(x)dx&=&\frac{\Big[\frac{1}{\epsilon^2}\big(2\cos(2\pi \epsilon\xi)-2\big)+4\pi^2\xi^2\Big]}{-4\pi^2\xi^2}\int\limits_{-\infty}^\infty e^{-2\pi i \xi x}f''(x)dx. \end{array}\end{equation*} Here we used integration by parts twice, which is allowed since the existence of the limits $\lim\limits_{x\rightarrow\pm\infty}f(x)$ and $\lim\limits_{x\rightarrow\pm\infty}f(x)’$ implies that $$\lim\limits_{x\rightarrow\pm\infty}\Delta_{\epsilon} f(x)=\lim\limits_{x\rightarrow\pm\infty}\Delta_{\epsilon} f'(x)=0.$$ Let us write $$q_\epsilon(\xi)=\frac{\frac{1}{\epsilon^2}\big(2\cos(2\pi \epsilon\xi)-2\big)+4\pi^2\xi^2}{-4\pi^2\xi^2},$$ so that the Fourier transform of $\Delta_\epsilon f-f''$ is the given by $q_\epsilon$ times the Fourier transform of $f''$. The goal is to estimate $||q_\epsilon \widehat{f''}||_{L^2(\mathbb{R})}$ and then apply Plancherel.
Since $-x^2\leq 2\cos(x)-2\leq 0$ always holds, we first note that $$q_\epsilon(\xi)=\frac{\frac{1}{\epsilon^2}\big(2\cos(2\pi \epsilon\xi)-2\big)-4\pi^2\xi^2}{-4\pi^2\xi^2}\leq \frac{-\frac{1}{\epsilon^2}4\pi^2\epsilon^2\xi^2+4\pi^2\xi^2}{-4\pi^2\xi^2} =0$$ and $$q_\epsilon(\xi)=\frac{\frac{1}{\epsilon^2}\big(2\cos(2\pi \epsilon\xi)-2\big)-4\pi^2\xi^2}{-4\pi^2\xi^2}\geq \frac{4\pi^2\xi^2}{-4\pi^2\xi^2}=-1,$$ which means that $q_\epsilon$ is nonnegative and bounded in $L^\infty(\mathbb{R})$ by $1$.
Now we pick $\delta>0$ and first note that since $f''\in L^2(\mathbb{R})$, we must have $\widehat{f''}\in L^2(\mathbb{R})$ as well. Now we pick $N>0$ in such a way that $$\int\limits_{(-\infty,-N]\cup[N,\infty)}|\widehat{f''}(\xi)|^2d\xi<\frac{\delta}{2},$$ which implies $$\int\limits_{(-\infty,-N]\cup[N,\infty)}|q_\epsilon(\xi)\widehat{f''}(\xi)|^2d\xi<\frac{\delta}{2},$$ since $|q_\epsilon|$ is bounded by $1$.
I claim that we can choose $\epsilon_0>0$ such that $|q_\epsilon(\xi)|\leq\frac{\delta}{2||\widehat{f''}||_{L^2(\mathbb{R})}^2}$ holds for all $0<\epsilon<\epsilon_0$ and all $-N\leq \xi\leq N$. Indeed, we let $K>0$ be the solution of the equation $$2-2\cos(x)=\left(1-\frac{\delta}{2||\widehat{f''}||_{L^2(\mathbb{R})}^2}\right)x^2,$$ which has a solution since we can assume without loss of generality that $0<\left(1-\frac{\delta}{2||\widehat{f''}||_{L^2(\mathbb{R})}^2}\right)<1$ and since $2-2\cos(x)=x^2+\mathcal{O}(x^4)$. Then, since $q_\epsilon\geq 0$, it follows that \begin{equation*} \begin{array}{llcl} &|q_\epsilon(\xi)|&\leq&\frac{\delta}{2||\widehat{f''}||_{L^2(\mathbb{R})}^2}\\ \Leftrightarrow&q_\epsilon(\xi)&\leq&\frac{\delta}{2||\widehat{f''}||_{L^2(\mathbb{R})}^2}\\ \Leftrightarrow& \frac{1}{\epsilon^2}\big(2\cos(2\pi \epsilon\xi)-2\big)&\leq&\left(\frac{\delta}{2||\widehat{f''}||_{L^2(\mathbb{R})}^2}-1\right)4\pi^2\xi^2\\ \Leftrightarrow&2-2\cos(2\pi \epsilon\xi)&\geq&\left(1-\frac{\delta}{2||\widehat{f''}||_{L^2(\mathbb{R})}^2}\right)(2\pi \epsilon\xi)^2\\ \Leftrightarrow&|2\pi \epsilon\xi|&\leq&K\\ \Leftrightarrow&|\xi|&\leq&\frac{K}{2\pi\epsilon}, \end{array} \end{equation*} which proves the claim by choosing $\epsilon_0$ as $N=\frac{K}{2\pi \epsilon_0}$. This way we can estimate, whenever $0<\epsilon<\epsilon_0$, $$\int\limits_{-N}^N|q_\epsilon(\xi)\widehat{f''}(\xi)|^2d\xi\leq \frac{\delta}{2||\widehat{f''}||_{L^2(\mathbb{R})}^2}\int\limits_{-N}^N|\widehat{f''}(\xi)|^2d\xi\leq \frac{\delta}{2}.$$
Finally, this shows that $$||q_\epsilon\widehat{f''}||_{L^2(\mathbb{R})}^2<\delta,$$ whenever $\epsilon<\epsilon_0$. By Plancherel, this yields that $\Delta_\epsilon f-f''\in L^2(\mathbb{R})$ and that $$\lim\limits_{\epsilon\downarrow 0}||\Delta_\epsilon f-f''||_{L^2(\mathbb{R})}=0,$$ as desired.