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Given $f$ is an even function of interval $(-a, a)$, $a>0$ and $0

$f$ is even then $f(-x)=f(x)$, then $f'(-x)=-f'(x)$ i.e $f'$ is odd function. But what exactly the question want? Please help with apropriate answer.

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    @user1942348: There is no assumption saying that $f$ is differentiable ... so you cannot simply write what you wrote under the colored frame. Instead, you have to *prove* that if $\lim_{h\to0^-}\frac{f(c+h)-f(c)}{h}$ exists (and is finite), then $\lim_{h\to0^+}\frac{f(-c+h)-f(-c)}{h}$ also exists and is the opposite of the previous limit. – 2017-02-20

1 Answers 1

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We have

\begin{align*} Lf'(c) &= \lim_{x\to c-} \frac{f(x)-f(x)}{x-c} = - \lim_{x\to c-} \frac{f(x)-f(x)}{c-x} \stackrel{1.}{=} - \lim_{x\to c+} \frac{f(-x)-f(c)}{c-(-x)} \\ & \stackrel{2.}{=} - \lim_{x\to c+} \frac{f(-x)-f(-c)}{c-(-x)} \stackrel{3.}{=} - \lim_{x\to c+} \frac{f(x)-f(-c)}{c-(-x)}=-Rf'(-c),\end{align*}

i.e. $$Lf'(c) = -Rf'(-c). \quad (\star)$$ Since $Lf'(c)$ exists, we know that $-Rf'(c)$ must exist aswell. In 1. we used the substitution $x\mapsto -x$ and in 2. and 3. we used that $f$ is even. If $f$ is differentiable in $(-a,a)$ then we know that $$f'(c)= Lf'(c) = Rf'(c) \quad \text{for all } c\in (-a,a).$$ Using $(\star)$ we get

$$f'(c)=-f'(-c) \quad \text{for all } c\in (-a,a),$$

i.e. $f'$ is an odd function on $(-a,a)$.

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    Please elaborate "Since Lf′(c) exists, we know that −Rf′(c) must exist aswell". – 2017-02-20