# Proving differentiability of a function with just epsilon deltas

Suppose f is a real-valued function on $$[−1, 1]$$

f has a continuous second derivative and $$f(0) = 0$$.

Define a function $$g$$ by $$g(x) = f'(0), x=0$$

$$g(x)=f(x)/x$$ at all other x.

Prove that g is continuous and differentiable at $$x = 0$$ and express $$g'(0)$$ in terms of the derivatives of $$f$$ at $$x = 0$$.

MY ATTEMPT:

Proving g is continuous was a matter of realizing it is equivalent to the differentiability of f at x=0. That we have, so we are done.

To know what $$g'(0)$$ should look like IF $$g$$ were differentiable, I used L'Hopital and got that it is $$-f''(0)/ 2$$

I am having trouble in using the epsilon delta definition of differentiability to prove that this is indeed the correct derivative. What I have at my disposable is : since second derivative exists, $$f'(x)$$ is differentiable and hence continuous. Similarly, f is continuous. But I can't seem to manipulate what I have to prove that this is the correct derivative. Can someone please help me?

• I missed something. Actually you can use L'Hopital because your assumption is that, $f$ has continuous second derivative around the point $0$. And you may look at my new answer, which is a bit complicated. – user284331 Oct 24 '19 at 5:24

\begin{align*} \left|\dfrac{g(h)-g(0)}{h}-\dfrac{1}{2}f''(0)\right|&=\left|\dfrac{1}{h^{2}}(f(h)-f(0)-f'(0)h)-\dfrac{1}{2}f''(0)\right|\\ &=\left|\dfrac{1}{h^{2}}\left(\int_{0}^{h}(f'(t)-f'(0))dt\right)-\dfrac{1}{2}f''(0)\right|\\ &=\left|\dfrac{1}{h^{2}}\int_{0}^{h}\int_{0}^{t}f''(u)dudt-\dfrac{1}{2}f''(0)\right|. \end{align*} For $$\epsilon>0$$, there is some $$\delta>0$$ such that $$|f''(u)-f''(0)|<\epsilon$$ for all $$|u|<\delta$$. Then for $$|h|<\delta$$, \begin{align*} f''(0)-\epsilon&\leq f''(u)\leq f''(0)+\epsilon\\ t(f''(0)-\epsilon)&\leq\int_{0}^{t}f''(u)du\leq t(f''(0)+\epsilon)\\ \int_{0}^{h}t(f''(0)-\epsilon)dt&\leq\int_{0}^{h}\int_{0}^{t}f''(u)dudt\leq\int_{0}^{h}t(f''(0)+\epsilon)dt\\ \dfrac{1}{2}h^{2}(f''(0)-\epsilon)&\leq\int_{0}^{h}\int_{0}^{t}f''(u)dudt\leq\dfrac{1}{2}h^{2}(f''(0)+\epsilon)\\ -\epsilon&\leq\dfrac{1}{h^{2}}\int_{0}^{h}\int_{0}^{t}f''(u)dudt-\dfrac{1}{2}f''(0)\leq\epsilon, \end{align*} so \begin{align*} \left|\dfrac{g(h)-g(0)}{h}-\dfrac{1}{2}f''(0)\right|\leq\epsilon. \end{align*}
• Last answer I made the mistake as you have pointed out too, that differential Mean Value gives you $\xi_{h}$ and $h$, they may not be equal, so the cancellation to $h^{2}$ is not allow. Sometimes switching to integrals can avoid this kind of problem. – user284331 Oct 24 '19 at 14:17