How to prove that $\displaystyle f''(a)=\lim_{h \to 0}\frac{f(a+2h)-2f(a+h)+f(a)}{h^2}$

Let $f:U\longrightarrow \mathbb{R}$ where $U\subset\mathbb{R}$ open set and $f$ twice-differentiable in $a\in U$.

How to prove that $$f''(a)=\lim_{h \to 0}\frac{f(a+2h)-2f(a+h)+f(a)}{h^2}$$

without using Taylor or L'Hopital's rule.

Any hints would be appreciated.

Since $f'(a) = \lim_{h\to 0} \frac{f(a + h) - f(a)}{h}$,

\begin{align*}f''(a) &= \lim_{h\to 0}\frac{ \lim_{m\to 0}\frac{ f(a + m + h) - f( a + h)}{m} - \lim_{m\to 0}\frac{f(a + m) - f(a)}{m}}{h} \\ &= \lim_{h\to 0}\lim_{m\to 0}\frac{f(a + m + h) - f(a + h) - f(a + m) + f(a)}{mh}. \end{align*}

This should look awfully suspicious. Since $m$ and $h$ are both going to zero, can you justify conflating them?

• [@Neal] Double limit exists ? $$\lim_{h\to 0 \\ m\to 0}\frac{f(a + m + h) - f(a + h) - f(a + m) + f(a)}{mh}$$ we can make $m=h$ ?. – felipeuni Sep 12 '13 at 3:23
• What you need to show here to make this rigorous is that if $f$ is twice differentiable, then the limit $\lim_{(h,m)\to (0,0)}\frac{(a+m+h)−f(a+h)−f(a+m)+f(a)}{mh}$ exists. This will allow you to conflate the two. – Baby Dragon Sep 12 '13 at 3:48
• my prof. suggested using taylor Polynomial but the problem is I understand you need to develop the Polynomial around two points (another hint i was given, otherwise you reach a dead end quiet quickly). since it's given that $f$ is only differentiable around the point $a$, i'm wondering what is the right way to approach this kind of method. – Jneven Oct 21 '18 at 9:03
• @Jneven I'd recommend asking that as its own question. :) – Neal Oct 22 '18 at 13:34

Using Barrow's rule: (assuming $f''$ is continuous at $x=a$) $$f(a+2h)-f(a+h) =\int_0^hf'(a+h+t)\,dt=\int_0^1 f'(a+h+sh)\,h\,ds$$ and

$$f(a+h)-f(a) =\int_0^hf'(a+t)\,dt=\int_0^1 f'(a+sh)\,h\,ds$$ Then $$\frac{f(a+2h)-2f(a+h)+f(a)}{h^2}=\frac{1}{h}\int_0^1 f'(a+h+sh) - f'(a+sh) \, ds \\=\frac{1}{h}\int_0^1\int_0^1f''(a+zh+sh)\,h\;dz\,ds \\ =\int_0^1\int_0^1f''(a+h(z+s))\;dzds \rightarrow f''(a) \mbox{ as } h\rightarrow 0$$

Without Taylor theorem or L'Hospital's Rule the problem essentially boils down to the proof of these two results. Here I use mean value theorem to prove Taylor's theorem and thereby the result in question.

Consider the function $$g$$ defined by $$g(x) =f(x) - f(a) - (x-a) f'(a) - \frac{(x-a) ^2}{2}f''(a)$$ then we have $$g(a)=g'(a)=g''(a)=0$$ By mean value theorem we have $$g(a+h) =g(a+h) - g(a)=hg'(a+\theta h)$$ where $$\theta\in(0,1)$$. Next note that by definition of derivative $$g'(a+\theta h) =g'(a) +\theta hg''(a) +o(\theta h) =o(h)$$ and then we have $$g(a+h) =o(h^2)$$ and therefore $$f(a+h) =f(a) +hf' (a) +\frac{h^2}{2}f''(a)+o(h^2)$$ Thus $$f(a+2h)-2f(a+h)+f(a)=h^2f''(a)+o(h^2)$$ and we are done.

I tried to develop Taylor polynomial of the second order around the point $$\ x=a\pm h\$$ , as $$\ h\to0$$, and to use Peano's remainder theorem

(Peano's remainder theorem: Let $$f:I\to\mathbb{R}$$ and some $$x_{0}\in I$$. Suppose $$f$$ is $$n$$- differentiable in the point $$x=x_{0}$$, then: $$R_{n}(x)=o\left(|x-x_{0}|^{n}\right)$$ as clearly $$\text{lim}_{x \to x_0} = 0$$ and that $$R_{n}(x_0) = 0$$. therefore this theorem allows you to divide $$R_n(x_0)$$ by $$(x-x_0)$$ for exactly n times without Without requiring additional properties on $$f$$)

Therefore:

$$f(a+h)=f(a)+f'(a)(h)+\frac{1}{2}f''(a)(h)^{2}+R_{2}(a+h)$$

$$f(a-h)=f(a)+f'(a)(-h)+\frac{1}{2}f''(a)(-h)^{2}+R_{2}(a-h)$$

$$\frac{R_{2}(a\pm h)}{h^{2}}\longrightarrow_{h\to0}=0$$

summing those equations, we'll get:

$$f(a+h)+f(a-h)=2f(a)+f'(a)h-f'(a)h+f''(a)h^{2}+R_{2}(a+h)+R_{2}(a-h)/-2f(a),\ \cdot\frac{1}{h^{2}}$$

$$\text{lim}_{h\to0}\frac{f(a+h)+f(a-h)-2f(a)}{h^{2}}=f''(a)$$