Suppose we have a function $f:[b,d]\to\mathbb{R}$ that has a jump discontinuity at some point $b<a<d$ and continuous otherwise. Define $F(x) = \int_b^x$ Then, from other answers on the site, I know that
- the jump discontinuity does not affect the value of $F(x)$
- $F(x)$ is continuous (since the jump discontinuity did not affect the value of $F(x)$, and $F(x)$ would be continuous/differentiable otherwise)
- $F(x)$ is not differentiable (at $a$)
When i think of a function not being differentiable I normally imagine a kink or a jump.
But I don't see how/why $F(x)$ would have a kink/jump (if that is the problem for differentiability). It can't jump because the discontinuity doesn't affect the value of the integral. For a kink I have no intuition
I tried computing an example where $$g:[-2,3]\to \mathbb{R} = \begin{cases} 1 & x\not = 0\\ 5 & x=0\end{cases}$$
Then we have, for$G(x)\equiv \int_{-2}^xg$ , that
$$
G(x) = \begin{cases} \int_{-2}^{x} 1 = (x) +2 =2+x & x< 0\\
\\lim_{\epsilon\to 0} \bigg(\int_{-2}^{-\epsilon}1ds + \int_{\epsilon}^x 1 ds = -\epsilon +2 + x-\epsilon = 2+x-2\epsilon\bigg ) = 2+x & x\geq 0 \end{cases}
$$
So the integral is $G(x) = 2+x$, which is continuous and differentiable. So I am messing up the calculation somewhere, but I don't see where? (perhaps I cannot write the integral as the limit as $\epsilon\to 0$, precisely because the discontinuity?)
Best Answer
In my mind "the" example of a kink is
$F(t) = 0$ for $t \leq 0$
$F(t) = t$ for $t \geq 0$.
This is zero for a while and then suddenly starts to increase. It is also clearly the integral of
$f(t) = 0$ for $t \leq 0$
$f(t) = 1$ for $t > 0$.
This is a source of kinks in higher derivatives too, because you can easily take e.g.
$G(t) = 0$ for $t \leq 0$
$G(t) = t^3$ for $t \geq 0$.
Clearly continuous. And 1st derivative is continuous. Second derivative is continuous but with a kink.