Let $f$ be a function defined on an interval $(a,b)$ and let $c \in (a,b)$. The symmetric derivative of $f$ at $c$ is defined by $f'_s(c)=lim_{h\to 0} \frac{f(c+h)-f(c-h)}{2h}$ provided that the limit exists. Prove that $f'_s(c)$ exists whenever $f'(c)$ exists, but that it is possible for $f'_s(c)$ to exists even when $f'(c)$ does not exist.

I have already proved it is possible for $f'_s(c)$ to exist when $f'(c)$ DNE, using the function $f(x)=|x|$.

But my struggle is with part 1: I know that we should assume that $f'(c)$ exists, and since $f'(c)$ exists then we know $lim_{x\to c} \frac{f(x)-f(c)}{x-c}$ exists. Now my struggle is getting from this to the definition of symmetric derivative.


\begin{eqnarray} \frac{f(x+h)-f(x-h)}{2h} &=& \frac{1}{2} \left( \frac{f(x+h)-f(x)+f(x)-f(x-h)}{h} \right) \\ &=& \frac{1}{2} \left( \frac{f(x+h)-f(x)}{h} + \frac{f(x)-f(x-h)}{h} \right) \\ &=& \frac{1}{2} \left( \frac{f(x+h)-f(x)}{h} + \frac{f(x-h)-f(x)}{-h} \right) \end{eqnarray} Now take limits.

Addendum: Since $\lim_{h \to 0} \frac{f(x+h)-f(x)}{h} = \lim_{h \to 0} \frac{f(x-h)-f(x)}{-h} = f'(x)$, we have that the corresponding limit on the left hand side exists and $\lim_{h \to 0} \frac{f(x+h)-f(x-h)}{2h} = f'(x)$.

  • $\begingroup$ I still don't understand what you are doing here, you are implying what I am trying to prove. $\endgroup$ – Arnold Nov 21 '13 at 20:55
  • $\begingroup$ The limit on the left is the symmetric difference, taking the limit as $h \to 0$ gives the symmetric derivative. Since it is equal to the right hand side, and both of those limits exist (and equal $f'(x)$) as $h \to 0$, then the limit on the left exists, which is what you were trying to prove. $\endgroup$ – copper.hat Nov 21 '13 at 21:05

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