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Let $f$ be a twice-differentiable function of $\mathbb R$ with $f(0)=0$. Define $$g(x)=\begin{cases}\frac{f(x)}{x},&\text{$x\neq0$}\\f'(0),&x=0\end{cases}$$ Prove that $g$ is a differentiable function of $x \in \mathbb R$.

I tried using the difference quotient around $0$ to get

$$g'(x)=\lim_{\epsilon \rightarrow 0} \frac{g(\epsilon)-g(0)}{\epsilon}=\lim_{\epsilon \rightarrow 0} \left( \frac{f(\epsilon)}{\epsilon^2} - \frac{f'(0)}{\epsilon}\right)$$ but this doesn't seem to be of much use. Apparently the problem can be solved using Taylor series, but I fail to see how.

actinidia
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4 Answers4

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$\lim_{x\to 0} \frac {f(x) -xf'(0)} {x^{2}} =\frac {f''(0)} 2$ by L'Hopital's Rule.

Kavi Rama Murthy
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  • @Botond Thanks. That was a typo. Corrected the answer. – Kavi Rama Murthy Nov 14 '18 at 09:52
  • Thank you! I thought that I did something wrong. – Botond Nov 14 '18 at 09:54
  • @KaviRamaMurthy Would you mind showing the calculations? For some reason L'Hôpital gave me $\frac{f'(x)-f'(0)}{2x}$. – actinidia Nov 14 '18 at 09:55
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    You have got it right up to this stage. Note that you still have $\frac 0 0$ form. If you apply L'Hopital's Rule again you get the answer. @TiwaAina – Kavi Rama Murthy Nov 14 '18 at 09:58
  • L'Hospital's Rule can be applied only once in a useful manner here. And then one can use definition of derivative to get $f''(0)/2$. – Paramanand Singh Nov 15 '18 at 14:50
  • @ParamanandSingh Applying the rule a second time is not necessary here, as you pointed out, but it is not wrong. Thanks anyway for the comment. – Kavi Rama Murthy Nov 15 '18 at 23:17
  • Well if you apply it second time then you get the expression $\lim_{x\to 0}f''(x)/2$. From the given condition of twice differentiability we can't conclude that this limit exists. We would need a hypothesis like $f$ is twice continuously differentiable. – Paramanand Singh Nov 16 '18 at 03:34
  • @ParamanandSingh You are right. I thought $f$ was a $C^{2}$ function when I wrote the comment. Thanks again for pointing out. – Kavi Rama Murthy Nov 16 '18 at 05:12
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By Taylor's theorem: $$f(x)=f(0)+f'(0)x+f''(0)\frac{x^2}{2}+r(x)x^2$$ $$f(x)=f'(0)x+\frac{f''(0)}{2}x^2+r(x)x^2$$ So $$\frac{f(x)}{x}=f'(0)+\frac{f''(0)}{2}x+r(x)x$$ So $g$ is continuous at $0$. Let's calculate $g'(0)$: $$g'(0)=\lim_{h \to 0} \frac{g(h)-g(0)}{h}$$ $$g'(0)=\lim_{h \to 0} \frac{\frac{f''(0)}{2}h+r(h)h}{h}$$ $$g'(0)=\lim_{h \to 0} \frac{f''(0)}{2}+r(h)$$ $$g'(0)=\frac{f''(0)}{2}$$

Botond
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  • Thank you! Is the observation that $g$ is continuous at $0$ especially important (for instance, if $g$ (or any other function) were somehow discontinuous at $0$, could it still be possible in theory to "find" a derivative at $0$ as we did? If so, would this derivative be extraneous?) – actinidia Nov 14 '18 at 09:53
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    $r(x)$ should be $r(h)$ in the computation for $g'(0)$. – egreg Nov 14 '18 at 10:07
  • @TiwaAina differentiability requires continuity. If $\lim_{h \to 0} g(x) \neq g(0)$, then the quotient is not in the $\frac{0}{0}$ form, but in the $\frac{something}{0}$ form, which is divergent. – Botond Nov 14 '18 at 10:12
  • Thank you @egreg! – Botond Nov 14 '18 at 10:13
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Use the Taylor development at $0$, $f(x)=xf'(0)+x\cdot \mathcal{O}(x)$ implies that

$\displaystyle{{g(x)-g(0)}\over{x}}={{f'(0)+\mathcal{O}(x)-f'(0)}\over x}$ where $\displaystyle\lim_{x\rightarrow 0}\mathcal{O}(x)=0$, this implies that $g$ is differentiable at $0$ and its differentiable on $\mathbb{R}$.

Yadati Kiran
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Tsemo Aristide
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First notice that $g$ is continuous at $0,$ by definition of $f'(0):=\lim_{x\to0}\frac{f(x)-f(0)}x.$

Moreover, as $x\to0,$ $$\begin{align}g'(x)&=\frac{xf'(x)-f(x)}{x^2}\\&=\frac{x\Big(f'(0)+xf''(0)+o(x)\Big)-\left(0+xf'(0)+\frac{x^2}2f''(0)+o(x^2)\right)}{x^2}\\&\to\frac{f''(0)}2. \end{align}$$

Therefore, $g$ is continuously differentiable at $0$ and $g'(0)=\frac{f''(0)}2.$

Anne Bauval
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  • Can the today-downvoter please explain how they believe I could improve my answer? (But this looks more like serial downvoting - 9 in two days - which will then be automatically reversed.) – Anne Bauval Sep 08 '23 at 13:41