Baby Rudin, Chapter 5, Exercise 11
Suppose $f$ is defined in a neighborhood of $x$, and suppose $f^{\prime\prime}(x)$ exists. Show that
\begin{equation}\tag{11.0}
\lim_{h \to 0} \frac{f(x+h)+ f(x-h)-2f(x)}{h^2} = f^{\prime\prime}(x)
\end{equation}
My attempt:
Firstly, notice that we can obtain alternative and equivalent versions of the definition of the derivative by making some notational maneuvers in the standard definition. We state the new definitions as follows: Let $f$ be defined (and real valued) on $[a, b]$. For any $x\in [a, b]$, the rate of change of the function $f$ at the point $x$, denoted by $f^{\prime}(x)$, is defined as
\begin{equation}\tag{11.1}
f^{\prime}(x) = \lim_{h \to 0}\; \frac{f(x)-f(x-h)}{x-(x-h)} = \lim_{h \to 0}\; \frac{f(x)-f(x-h)}{h}
\end{equation}
where $a< t< b$ and $t \ne x$. Moreover, leaving everything else unchanged, we can rewrite $(11.1)$ as
\begin{equation}\tag{11.2}
f^{\prime}(x) = \lim_{h \to 0}\; \frac{f(x+h)-f(x)}{x+h-(x)} = \lim_{h \to 0}\; \frac{f(x+h)-f(x)}{h}
\end{equation}
Next, if we assume that $f^{\prime}(x)$ exists in a neighborhood of $x$ and $f^{\prime}$ is differentiable at the point $x$, then, using (11.2) we can define $f^{\prime\prime}(x)$ as:
\begin{equation}\tag{11.3}
f^{\prime\prime}(x) = \lim_{h \to 0}\; \frac{f^{\prime}(x+h)-f^{\prime}(x)}{h}
\end{equation}We only need to perform some routine algebra to show (11.0). Suppose $f$ is defined in a neighborhood of $x, [a, b]$, and suppose $f^{\prime\prime}(x)$ exists. Then, we know that $f^\prime$ exists in a neighborhood of $x$ and is differentiable at $x$. Thus, (11.3) holds and by (11.1), we have
\begin{equation}\tag{11.4}
f^{\prime}(x+h) = \lim_{h \to 0}\; \frac{f(x+h)-f[(x+h)-h]}{(x+h)-[(x+h)-h]} = \lim_{h \to 0}\; \frac{f(x+h)-f(x)}{h}
\end{equation}
By substituting (11.4) and (11.1) in (11.3), we get
\begin{align*}
f^{\prime\prime}(x) &= \frac{1}{h}\cdot \lim_{h \to 0}\; \left[\frac{f(x+h)-f(x)}{h} – \left(\frac{f(x)-f(x-h)}{h}\right)\right] \\
&= \frac{1}{h^2} \lim_{h \to 0}\; \left[f(x+h)-f(x)-f(x)+f(x-h)\right]
\end{align*}
which proves (11.0).
My question: Is my proof correct? If not, how can the mistakes in this proof be patched up? Especially, is the expression in (11.4) and the process for obtaining (11.4) correct? I've basically shown that the right hand sides of (11.4) and (11.2) are the same, despite the fact their left hand sided are not equal.
Best Answer
Hope you found your mistakes as @Qiyu Wen pointed out. Alternatively, you can prove easily by "L-Hospital" Rule:
Hints: \begin{align*} f''(x)=\operatorname{lim}_{h\to 0}\Big[\frac{f(x+h)-2f(x)+f(x-h)}{h^2}\Big] =\operatorname{lim}_{h\to 0}\Big[\frac{f'(x+h)+f'(x-h)}{2h}\Big] =\frac{1}{2}\operatorname{lim}_{h\to 0}\Big[\frac{f'(x+h)-f'(x)}{h}+\frac{f'(x)-f'(x-h)}{h}\Big] \end{align*}