Suppose that $f''(0)$ exists and is finite, show that $f''(0)=\lim\limits_{h \to 0}\frac{f(h)-2f(0)+f(-h)}{h^{2}}$

I read the following from Chung's Probability:

($$f$$ is the characteristic function of some distribution function $$F$$) Suppose that $$f''(0)$$ exists and is finite, then we have $$f''(0)=\lim\limits_{h \to 0}\frac{f(h)-2f(0)+f(-h)}{h^{2}}$$

Could anyone help on how to show this? The problem is that we only assume $$f$$ is twice differentiable at the point 0. So we can't prove the above simply by Taylor's expansion.

Thank you very much!

• I could be misunderstanding the question, but isn't this just the definition of the second derivative? – icurays1 Nov 12 '12 at 19:28
• I suggest writing down the taylor expansion of $f(x),f(-x)$, if I recall correctly this should do most of the work. – Belgi Nov 12 '12 at 19:38
• @icurays1 No, second derivative is the derivative of the derivative. This is a formula often used in numerical approximation, but it is not the definition. – Arthur Nov 12 '12 at 19:39
• Am I missing something, or is it possible to just apply L'Hôpital's rule twice to the right side limit? – yunone Nov 12 '12 at 19:47
• If you write down what it means to be the 'derivative of the derivative' using the difference quotient/limit definition of $f^\prime$, you arrive at this limit definition for $f^{\prime\prime}$. So, you could take it as an equivalent definition if you wanted. – icurays1 Nov 12 '12 at 19:47

Taylor's theorem says that if $f$ is twice differentiable at $0$, $f(x) = f(0) + f'(0) x + f''(0) \frac{x^2}{2} + R(x) x^2$, where $R(x) \to 0$ as $x \to 0$. Apply this at $x=h$ and $x=-h$ and substitute: $$\dfrac{f(h)- 2f(0) + f(-h)}{h^2} = f''(0) + R(h) + R(-h) \to f''(0)\ \text{as }\ h \to 0$$