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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?

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  • $\begingroup$ 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. $\endgroup$ – user284331 Oct 24 '19 at 5:24
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\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*}

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  • $\begingroup$ may I ask what the motivation was to use FTC instead of trying to prove this by brute force? Is it because you saw that you had to relate the function with its first and second derivatives and the question gave us some nice properties for these derivatives? $\endgroup$ – childishsadbino Oct 24 '19 at 13:29
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    $\begingroup$ 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. $\endgroup$ – user284331 Oct 24 '19 at 14:17

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