Show $\lim_{h \to \ 0} \frac{f(x + 2h) - 2f(x+h) + f(x)}{h^{2}} = f''(x)$ Proof verification

Solution 1:

We have $$f^{\prime}(x+h)-f^{\prime}(x)=\lim_{k\to0}\frac{f(x+h+k)-f(x+h)}{k}-\lim_{k\to0}\frac{f(x+k)-f(x)}{k}\\=\lim_{k\to0}\frac{f(x+h+k)-f(x+h)-f(x+k)+f(x)}{k}$$and so$$f^{\prime\prime}(x)=\lim_{h\to0}\lim_{k\to0}\frac{f(x+h+k)-f(x+h)-f(x+k)+f(x)}{kh}.$$You're right to feel "discomfort": we need to justify how we go from using two distinct tend-to-$0$ variables to just one. Here's my understanding (but a greater expert on analysis might say this isn't the right way to do it):

The quantity whose double limit is taken is $h\leftrightarrow k$-symmetric, so iff $f^{\prime\prime}(x)$ exists we can unambiguously write$$f^{\prime\prime}(x)=\lim_{h,\,k\to0}\frac{f(x+h+k)-f(x+h)-f(x+k)+f(x)}{kh}.$$Then we can take $h,\,k$ to $0$ in any way we like, all giving the same result, so let's take $k=h$. Then$$f^{\prime\prime}(x)=\lim_{h\to0}\frac{f(x+2h)-2f(x+h)+f(x)}{h^2}.$$

Edit: taking advice from comments @ParamanandSingh left under the OP and this answer, let's do it in a less suspect way, which does use just one tending-to-$0$ variable but not in the way the OP tried. Using L'Hôpital's rule,$$\lim_{h\to 0}\frac{f(x+2h)-2f(x+h)+f(x)}{h^2}=\lim_{h\to 0}\frac{f^\prime(x+2h)-f^\prime(x+h)}{h}\\=2\lim_{h\to 0}\frac{f^\prime(x+2h)-f^\prime(x)}{2h}-\lim_{h\to 0}\frac{f^\prime(x+h)-f^\prime(x)}{h}=2f^{\prime\prime}(x)-f^{\prime\prime}(x)=f^{\prime\prime}(x).$$

Solution 2:

Let $f$ be any continuous function which is second differentiable at $a$ (i.e. $f''(a)$ exists). By definition of second differentiability at a point, $f'(x)$ exists over $(a-\eta,a+\eta)$ for some $\eta > 0$. Define $g : (-\eta, \eta) \to \mathbb{R}$ by

$$g(x) = f(a+x) - f(a) - f'(a) x - \frac12 f''(a) x^2$$

It is clear $g$ is continuous, second differentiable at $a$ and $g'(x)$ exists over $(-\eta,\eta)$. Furthermore, $g(0) = g'(0) = g''(0) = 0$. What we want to show is equivalent to following statement

$$\lim_{h\to 0}\frac{g(2h) - 2g(h) + g(0)}{h^2} = 0 \tag{*1}$$

Since $f''(a)$ exists, $g''(0) = \lim_{h\to 0}\frac{g'(h)}{h}$ exists and vanishes. For any $\epsilon > 0$, there exists a $\delta \in ( 0,\eta)$ such that

$$\left|\frac{g'(x)}{x}\right| = \left|\frac{g'(x) - g'(0)}{x}\right| < \frac{\epsilon}{3}\quad\text{ whenever }\quad 0 < |x| < \delta$$

For any $h$ such that $0 < |h| < \frac{\delta}{2}$, apply MVT to $g(2h) - g(h)$ and $g(h) - g(0)$, we find there are $p,q \in (0,1)$ such that $$\frac{g(2h) - g(h)}{h} = g'((1+p)h)\quad\text{ and }\quad \frac{g(h) - g(0)}{h} = g'(qh)$$

This leads to

$$\begin{align}\left|\frac{g(2h) - 2g(h) + g(0)}{h^2}\right| & \le \left| \frac{g(2h) - g(h)}{h^2}\right| + \left|\frac{g(h) - g(0)}{h^2}\right|\\ &= \left|\frac{g'((1+p)h)}{h}\right| + \left|\frac{g'(qh)}{h}\right| \\ & < \frac{\epsilon}{3}\left((1+p) + q\right)\\ & < \epsilon \end{align} $$

Since $\epsilon$ can be arbitrary small, $(*1)$ follows and hence $$\lim_{h\to 0}\frac{f(a+2h) - 2f(a+h) + f(a)}{h^2} = f''(a)$$