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If $ f $ is a continuous function defined on a real interval that has a discontinuity at a point (but is continuous otherwise), then is it possible to be differentiable at that point?

user42
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  • Can you please define simple discontinuity? – snar Mar 09 '15 at 17:57
  • yes simple discontinuity is a point at which left and right hand limits exist; but either don't equal each other or don't equal the value of the function at that point – user42 Mar 09 '15 at 18:01
  • Even if $f$ has a "non-simple" discontinuity, for example, $f$ is a step function then $\int_0^x f(y)dy$ is differentiable but the derivative is not continuous of course. – Martingalo Mar 09 '15 at 18:04
  • Thank you I was misunderstanding something – user42 Mar 09 '15 at 18:07
  • The answer to your question is simply "no". Differentiability is a stronger condition than continuity; moreover, differentiability at a point implies continuity at that point. Said differently, if $f$ is not continuous at a point $x$, then $f$ cannot be differentiable at $x$. – MPW Mar 09 '15 at 18:39

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Let $f$ be a continuous function defined on $[0,1]$, except at $t = 1/2$ where both limits $f(x-)$ and $f(x+)$ exist: $$\lim_{x \rightarrow t,\ x < t} f(x) = f(x-), \quad \lim_{x \rightarrow t,\ x > t} = f(x+).$$ It is possible that $F(x) = \int_0^x f(t) dt$ is differentiable at $x = 1/2$ and it is possible that $F(x)$ is not differentiable at $x = 1/2$.

For the latter case, consider $f(t) = 1_{[1/2,1]}(t)$, the indicator function on $[1/2, 1]$. Then $F(x) = (x-1/2)1_{[1/2,1]}(x)$ and $F$ is not differentiable at $x$ as $\frac{F(x+h)-F(x)}{h} \rightarrow 1$ from the right and 0 from the left as $h \rightarrow 0$.

For a positive result, consider $$f(t) = \begin{cases} 0 & t \neq 1/2 \\ 1 & t = 1/2\end{cases}.$$ Then the Riemann-Stieltjes integral $F(x)$ exists and is equal to 0, which is differentiable on all of $[0,1]$, in particular at $ x = 1/2.$

snar
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