# Spivak: Determine if a function $F(x)=\int_0^x f$ is differentiable at $0$. $f$ is a function given by a particular graph.

The following is a problem from Ch. 14 "The Fundamental Theorem of Calculus", from Spivak's Calculus

Let $$F(x)=\int_0^x f$$.

Consider the function depicted in the picture below

At which points $$x$$ is $$F'(x)=f(x)$$?

The solution manual says

All $$x\neq 0$$. $$F$$ is not differentiable at $$0$$ because $$F(x)=0$$ for $$x\leq 0$$, but there are $$x>0$$ arbitrarily close to 0 with $$\frac{F(x)}{x}=\frac{1}{2}$$

I'd like to understand this solution better. Here is my attempt at filling in the intermediate steps in the proof.

$$f$$ is continuous everywhere except at $$x=0$$. Thus, we can apply the first fundamental theorem of calculus to conclude that

$$F'(x)=f(x)$$

when $$x\neq 0$$.

$$f$$ also happens to be integrable everywhere.

What happens at $$x=0$$ is the question.

If $$F$$ is differentiable at $$0$$ then the following limits exist and are equal to each other

$$\lim\limits_{h\to 0^-} \frac{F(h)}{h}=\lim\limits_{h\to 0^+} \frac{F(h)}{h}$$

Now

$$\lim\limits_{h\to 0^-} \frac{F(h)}{h}=0$$

Do we also have $$\lim\limits_{h\to 0^+} \frac{F(h)}{h}=0$$?

There exists an $$n\in\mathbb{N}$$ such that $$\frac{1}{2^n}.

The sum of areas of triangles formed by $$f$$ up to $$\frac{1}{2^n}$$ is

$$F(1/2^n)=\frac{1 \cdot \sum\limits_{i=n}^\infty \left ( \frac{1}{2^i}-\frac{1}{2^{i+1}}\right )}{2}$$

I'm not sure what the infinite sum is, but I'm guessing it is $$\frac{1}{2^n}$$

Thus

$$\frac{F(1/2^n)}{2^n}=\frac{1}{2}$$

That is, there is always $$x$$ such that $$0 and

$$\frac{F(x)}{x}=\frac{1}{2}$$

Therefore, $$\lim\limits_{h\to 0^+} \frac{F(h)}{h}$$ doesn't exist.

Hence $$F$$ is not differentiable at $$0$$.

Is this correct?

Pretty much. There are a couple of little errors that can be corrected easily enough. They are just typos, stylistic errors, and small oversights, so I do think your intuition is perfectly fine.

First, there's what I think is a typo:

$$\frac{F(1/2^n)}{2^n}=\frac{1}{2}$$

should be

$$\frac{F(1/2^n)}{1/2^n}=\frac{1}{2}.$$

Second, you should introduce $$h$$ before you write:

There exists an $$n\in\mathbb{N}$$ such that $$\frac{1}{2^n}.

Before this, it was a dummy variable. What you should write instead is,

Given any $$h > 0$$, there exists an $$n\in\mathbb{N}$$ such that $$\frac{1}{2^n}.

Third, the fact that $$F(x) / x = 1/2$$, for our constructed $$x \in (0, h)$$, does not quite imply that $$\lim_{h \to 0^+} F(h) / h$$ does not exist! Instead, it proves that if it exists, it must be $$1/2$$. This is sufficient, however, because $$\lim_{h \to 0^-} F(h) / h$$ is $$0$$, as you showed. If the two-sided limit existed, the one-sided limits would have to both exist and agree, which would make $$\lim_{h \to 0^+} F(h) / h = 0$$. Your argument proves this is impossible.

Finally, I thought I'd point out why the sum is $$\frac{1}{2^n}$$. The series you're looking at here is telescoping! It is geometric too, but let's do it with telescoping series. When we expand any partial sum, most of the terms cancel:

\begin{align*} \sum_{i=n}^N \frac{1}{2^i} - \frac{1}{2^{i+1}} &= \frac{1}{2^n} - \frac{1}{2^{n+1}} + \frac{1}{2^{n+1}} - \frac{1}{2^{n+2}} + \frac{1}{2^{n+2}} - \frac{1}{2^{n+3}} + \ldots + \frac{1}{2^N} - \frac{1}{2^{N+1}} \\ &= \frac{1}{2^n} - \frac{1}{2^{N+1}} \end{align*}

As $$N \to \infty$$, this approaches $$\frac{1}{2^n}$$.

Take the sequence $$\,y_{n}$$ of points such that $$f(y_{n})=1$$. It turns out that

$$y_{n}=3/2^{n+1}$$. Now consider $$\int_{0}^{y_{n}}f(t)dt.$$ It is easy to check that

$$\int_{0}^{y_{1}}f(t)dt=\frac{1}{8}(1+\frac{1}{2}+\ldots)=\frac{1}{4}.\,\,$$ Likewise

$$\int_{0}^{y_{2}}f(t)dt=\frac{1}{16}(1+\frac{1}{2}+\ldots)=\frac{1}{8}$$ and $$\,\,\int_{0}^{y_{n}}f(t)dt=1/2^{n+1}$$.

If we consider the ratio $$\int_{0}^{y_{n}}f(t)dt/y_{n}$$ we see

that is $$1/3$$ and hence the limit is $$1/3$$ which implies that

lim$$(F(y_{n})/y_{n})$$ is NOT zero as it should be if $$F$$ was differentiable at $$x=0$$.

Therefore the function $$F$$ is not differentiable at $$x=0$$!

• L'Hopital's rule doesn't work this way. L'Hopital's rule says that, if $\lim \frac{f'(x)}{g'(x)}$ exists and is equal to $L$, then $\lim \frac{f(x)}{g(x)}$ exists and is equal to $L$. The inverse implication doesn't hold in general, in that if $\lim \frac{f'(x)}{g'(x)}$ does not exist, then $\lim \frac{f(x)}{g(x)}$ does not exist. Jul 1, 2022 at 23:34
• I agree! Thanks for correcting that!
– user1054388
Jul 2, 2022 at 0:54
• I appreciate you trying something new here, but unfortunately, I also have qualms with this application of Stolz-Cesaro. In order for Stolz-Cesaro to apply, the sequence in the denominator (denoted $(b_n)$ in the wiki article) must be strictly monotone and divergent. In this case, $y_n$ is strictly monotone, but convergent to $0$. Jul 3, 2022 at 1:06