Is Spivak wrong about this counterexample? $f$ integrable on $[-1,1]$, $F=\int_{-1}^xf$, $f$ differentiable at $0$, but $F'$ not continuous at $0$

Solution 1:

The answer to your question in the comments is yes: If $f$ is differentiable at $c,$ then $F'$ is continuous at $c$ within the domain of $F'.$

Proof: Suppose $y_n\to c^+$ (the case $y_n\to c^-$ is similar). Assume $F'(y_n)$ exists for each $n.$ Because $f'(c)$ exists, we can write

$$\tag 1 f(x) = f(c) +f'(c)(x-c) + \epsilon(x)(x-c),$$

where $\epsilon(x)\to 0$ as $x\to c.$ Actually we don't need the $\epsilon\to 0$ property. If you divide both sides of $(1)$ by $x-c,$ you'll see $\epsilon(x)$ is bounded on $[a,b]\setminus \{c\},$ say by $M$ in absolute value.

Fix $n.$ Then for small $h>0,$

$$\tag 2 \frac{F(y_n+h)-F(y_n)}{h} = \frac{1}{h}\int_{y_n}^{y_n+h} f(x)\,dx.$$

By $(1),$ $(2)$ equals

$$\tag 3 f(c) + \frac{1}{h}\int_{y_n}^{y_n+h}(f'(c)+\epsilon(x))(x-c)\,dx.$$

Now the second expression in $(3)$ is bounded above in absolute value by

$$(|f'(c)|+M)\frac{1}{h}\int_{y_n}^{y_n+h}(x-c)\,dx = (|f'(c)|+M)(y_n-c+h).$$

Thus

$$\left |\frac{F(y_n+h)-F(y_n)}{h} - f(c)\right |\le (|f'(c)|+M)(y_n-c+h).$$

Taking the limit as $h\to 0$ then gives

$$|F'(y_n) - f(c)| \le (|f'(c)|+M)(y_n-c).$$

Now letting $n\to \infty,$ we see

$$F'(y_n) \to f(c) = F'(c),$$

and we're done.