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I do not have mean value theorem and can't use it. All I have are IVT and that the derivative of an even function is odd and the definition of a derivative. With those tools, what can I do?

cakey
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    You mean an even function which is differentiable at $x=0$? Clearly derivative of $|x|$ at $x=0$ DNE. – MonkeyKing Apr 23 '15 at 15:14
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    You must be given much more, otherwise the claim fails big time even with a continuous function, as $;f(x)=|x|;$ shows... – Timbuc Apr 23 '15 at 15:14
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    Think about $f(0-)$, and $f(0+)$. If these coincide, and they are both equal to $f(0)$ then the function is continuous at $0$. – Diego Robayo Apr 23 '15 at 15:27

3 Answers3

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Suppose you mean the even function is differentiable at $x=0$, then use the definition. The left derivative is:

$$\lim_{h\rightarrow 0^+}\frac{f(0)-f(-h)}{h}=\lim_{h\rightarrow 0^+}-\frac{f(-h)-f(0)}{h}$$

The right derivative is: $$\lim_{h\rightarrow 0^+}\frac{f(h)-f(0)}{h}$$

Since $f$ is even, then $f(-h)=f(h)$. The left derivative must equal to right derivative, so they have to be $0$.

MonkeyKing
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    So I understand making the left derivative and right derivatives equal, I'd end up with -f(h)=f(h) which is only true when f(h)=0 and ultimately reach equality of the derivatives which is only true when the derivative is zero. How did you know for the left to make f(0) come first?As in before f(-h)? And how do I get that it is negative h intuitively? – cakey Apr 23 '15 at 15:36
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    @cakey Well, you just need to name the right derivative something like $f'$. The manipulation above just makes you observe that left derivative is $-f'$, and moreover, we have $f'=-f'$. – MonkeyKing Apr 23 '15 at 15:41
  • There are a few implicit steps in here that (at least for me) make the proof more clear. By assumption, we have that $\displaystyle \lim_{h \to 0^-}\frac{f(h)-f(0)}{h}=\displaystyle \lim_{h \to 0^+}\frac{f(h)-f(0)}{h}=M$. Because $\displaystyle \lim_{h \to 0^-}\frac{f(h)-f(0)}{h}$ exists, we know that the following limit exists, too: $\displaystyle \lim_{h \to 0^-}(-1)\cdot\frac{f(h)-f(0)}{h}=-M$. – S.C. Oct 20 '22 at 20:30
  • Because $f$ is an even function, we can rewrite and further simplify the function $(-1)\cdot \frac{f(h)-f(0)}{h}$ as $\frac{f(-h)-f(0)}{-h}$. Note, in this left limit context, $h \lt 0$, which means that $-h \gt 0$. Therefore, we see that $\displaystyle \lim_{h \to 0^-}(-1)\cdot\frac{f(h)-f(0)}{h}$ contains the same information as $\displaystyle \lim_{h \to 0^+}\frac{f(h)-f(0)}{h}$. A such, we note that $-M=\displaystyle \lim_{h \to 0^-}(-1)\cdot\frac{f(h)-f(0)}{h}=\lim_{h \to 0^+}\frac{f(h)-f(0)}{h}=M$. Under what circumstances does $-M=M$? Precisely and only when $M=0$. – S.C. Oct 20 '22 at 20:30
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Hint: What do you know about an odd function evaluated at $x=0$?

Hagen von Eitzen
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I could not understand the accepted answer, so here is another proof of the following statement:

If $f$ is an even function that is differentiable at $x=0$, then $f'(0)=0$.

Proof

By assumption, we know that $\displaystyle \lim_{x \to 0} \frac{f(x)-f(0)}{x-0}=M$. WLOG, suppose $M \gt 0$. Now, consider an $0\lt\varepsilon \lt \frac{M}{2}$. Then there is a corresponding $\delta_{\varepsilon} \gt 0: \forall x \in (0-\delta_{\varepsilon}, 0+\delta_{\varepsilon})\setminus\{0\}:\left|\frac{f(x)-f(0)}{x-0} -M\right| \lt \varepsilon \quad (\dagger_1)$.

Consider a $z_L$ and a $z_R$ such that $z_R \in (0,\delta_{\varepsilon})$ and $z_L=-z_R$. By $(\dagger_1)$, we know the following:

  • $\left|\frac{f(z_R)-f(0)}{z_R}-M \right| \lt \varepsilon \implies M-\varepsilon \lt \frac{f(z_R)-f(0)}{z_R} \lt M+\varepsilon \quad (\dagger_2)$

  • $\left|\frac{f(z_L)-f(0)}{z_L}-M \right| \lt \varepsilon \implies M-\varepsilon \lt \frac{f(z_L)-f(0)}{z_L} \lt M+\varepsilon \quad (\dagger_3)$

Because $f$ is an even function, which means that $\forall x \in \text{dom}(f): f(x)=f(-x)$, and we designated that $z_L=-z_R$, we can rewrite $(\dagger_3)$ in the following way:

$(\dagger_3) \iff M-\varepsilon \lt \frac{f(z_R)-f(0)}{-z_R} \lt M +\varepsilon$.

Multiplying the inequality by $(-1)$ gives us:

  • $-M+\varepsilon \gt \frac{f(z_R)-f(0)}{z_R} \gt -M -\varepsilon \quad (\dagger_4)$

Because $M \gt 0$ and $0 \lt \varepsilon \lt \frac{M}{2}$, we know that $-M + \varepsilon \lt 0$ and $M-\varepsilon \gt 0$. Under these circumstances, $(\dagger_2)$ and $(\dagger_4)$ can be combined to give:

$$-M-\varepsilon \lt \frac{f(z_R)-f(0)}{z_R} \lt -M+\varepsilon \lt 0 \lt M-\varepsilon \lt \frac{f(z_R)-f(0)}{z_R} \lt M+\varepsilon$$

which, yields the contradictory statement of: $\frac{f(z_R)-f(0)}{z_R} \lt \frac{f(z_R)-f(0)}{z_R}$.

A similar contradiction will arise if $M \lt 0$.

We therefore conclude that $M=0$.

S.C.
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