# Can someone check my proof for non-differentiability?

• Jul 17th 2013, 12:26 PM
nimon
Can someone check my proof for non-differentiability?
Hi all. Doing a problem led me to derive this little lemma, but I'd like someone to check it for me if you would be so kind.

Theorem: Let $\displaystyle f$ be an integrable function and define $\displaystyle F(x) = \int_{0}^{x} f(t)\,dt.$ Suppose that, for some $\displaystyle c$ in the domain of $\displaystyle f$ we have $\displaystyle \lim\limits_{x \rightarrow c^{-}} f(x) \neq \lim\limits_{x\rightarrow c^{+}} f(x),$ where the limits are presumed to exist. Then $\displaystyle F(x)$ is not differentiable at $\displaystyle c.$

Proof: Assume without loss of generality that $\displaystyle \lim\limits_{x \rightarrow c^{-}} f(x) = \ell_{1} < \ell_{2} \lim\limits_{x\rightarrow c^{+}} f(x).$ Then there is some $\displaystyle \delta>0$ such that $\displaystyle 0 < |x-a| < \delta$ implies:
1. $\displaystyle f(x) < \ell_{1} + \frac{\ell_{2}-\ell_{1}}{4} = m$ for $\displaystyle x<c$
2. $\displaystyle f(x) < \ell_{2} - \frac{\ell_{2}-\ell_{1}}{4} = M$ for $\displaystyle x>c$

Clearly $\displaystyle M>m.$ Then for any $\displaystyle 0<h<\delta$ we have

$\displaystyle \frac{F(x+h) - F(x)}{h} = \frac{1}{h} \int_{x}^{x+h} f(t)\,dt > \frac{Mh}{h} = M$
and for any $\displaystyle -\delta < h < 0$ we have

$\displaystyle \frac{F(x+h) - F(x)}{h} = -\frac{1}{h} \int_{x+h}^{x} f(t)\,dt < -\frac{1}{h} \cdot -mh = m$
so that $\displaystyle \lim\limits_{h\rightarrow 0} \frac{F(x+h)-F(x)}{h}$ cannot exist. QED.

Any help is much appreciated!