Principles of Mathematical Analysis, Rudin, 3th ed, Chapter 5, Problem 11

Problem

Suppose $f$ is defined in a neighborhood of $x$, and suppose $f''(x)$ exists. Show that

$$ \lim_{h \rightarrow 0} \frac{f(x+h)+f(x-h)-2f(x)}{h^2} = f''(x).$$

Show by an example that the limit may exists even if $f''(x)$ does not.

Answer

Since $f''$ exists at $x$, $f'$ exists in a neighborhood of $x$.

$\lim_{h \rightarrow 0} \frac{f(x+h)+f(x-h)-2f(x)}{h^2} = \lim_{h \rightarrow 0} \frac{\frac{d}{dh}(f(x+h)+f(x-h)-2f(x))}{\frac{d}{dh}h^2} = \lim_{h \rightarrow 0}\frac{f'(x+h)-f'(x-h)}{2h} = \lim_{h \rightarrow 0} \frac{f'(x+h)-f'(x)}{2h} + \lim_{h \rightarrow 0} \frac{f'(x-h)-f'(x)}{-2h} = f''(x).$

If $f(x)=x|x|$, $f''(0)$ does not exist, but

$$ \lim_{h \rightarrow 0}\frac{f(0+h) + f(0-h) - 2f(0)}{h^2} - \lim_{h \rightarrow 0} \frac{h|h| - h|h|}{h^2} =0.$$ 

Remark: If $f''$ exists on the domain of $f$, then we have another way to prove the proposition; there exist $c_1 \in (x,x+h),c_2 \in (x-h,x)$ such that $f(x+h) = f(x) + f'(x)h + f''(c_1)h^2/2$, $f(x-h) = f(x) - f(x)h + f''(c_2)h^2/2$ by Theorem 5.15. Then $\lim_{h \rightarrow 0} \frac{f(x+h)+f(x-h)-2f(x)}{h^2} = \lim_{h \rightarrow 0} \frac{f''(c_1)h^2/2 + f''(c_2)h^2/2}{h^2} = f''(x).$

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