Is this form of Taylor's theorem correct? Güler's Foundations of Optimization theorem 1.3

Question

Let $f: I = (c, d) \xrightarrow{} \mathbb{R}$ be an $n-$times differentiable function. Let $a, b$ be distinct points in $I$. Then

$f(b) = f(a) + f'(a)(b-a) + ... + \frac{f^{(n-1)}(a)}{(n-1)!}(b-a)^{n-1} + \int_{a}^{b} \int_{a}^{s_{1}} ... \int_{a}^{s_{n-1}} f^{(n)}(s_{n}) ds_{n}...ds_{1}.$

Proof. By the Fundamental Thm of Calculus, we have $f(b) = f(a) + \int_{a}^{b}f'(s_{1})ds_{1}$. Similarly, $f'(s_{1}) = f'(a) + \int_{a}^{s_{1}}f''(s_{2})ds_{2}$, which gives

$f(b) = f(a) + \int_{a}^{b}f'(s_{1})ds_{1} = f(a) + \int_{a}^{b}(f'(a) + \int_{a}^{s_{1}}f''(s_{2})ds_{2})ds_{1} = f(a) + f'(a)(b-a) + \int_{a}^{b} \int_{a}^{s_{1}} f''(s_{2}) ds_{2} ds_{1}.$

Continuing this fashion, we obtain $f(b) = f(a) + f'(a)(b-a) + f''(a)\int_{a}^{b} \int_{a}^{s_{1}} \int_{a}^{s_{2}} ds_{3}ds_{2}ds_{1} +...+ f^{(n-1)}(a)\int_{a}^{b}\int_{a}^{s_{1}} ... \int_{a}^{s_{n-1}}ds_{n}...ds_{1} + \int_{a}^{b}\int_{a}^{s_{1}}...\int_{a}^{s_{n-1}}f^{(n)}(s_{n})ds_{n}...ds_{1}.$

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My question is, in the last step the author obviously used $f^{(n-1)}(s_{n-1}) - f^{(n-1)}(a) = \int_{a}^{s_{n-1}}f^{(n)}(s_{n})ds_{n}$, but why is this true? The Fundamental theorem says if $f$ is integrable on the interval $[a, b]$ and $F' = f$ there, then $\int_{a}^{b} f dx = F(b) - F(a)$. But here, $f^{(n)}$ may not be integrable, so how do we get $f^{(n-1)}(s_{n-1}) - f^{(n-1)}(a) = \int_{a}^{s_{n-1}}f^{(n)}(s_{n})ds_{n}$?

Should we assume $f^{(n)}$ is continuous or at least Riemann integrable to apply the Fundamental Thm of Calculus?

The book can be downloaded here https://link.springer.com/book/10.1007/978-0-387-68407-9

Answer 1

I suspect the author of this book may be sloppy with hypotheses.

Yes, it is standard to state such results with the assumption that $f\in C^n([a,b])$.

You can certainly give counterexamples. It's easiest to give the counterexample for $n=1$. If you take $$f(x) = \begin{cases} x^2\sin(1/x^2), & x\ne 0 \\ 0, & x=0\end{cases},$$ then $f$ is differentiable but $f'$ is unbounded on $[0,1]$, and so the Riemann integral of $f'$ is not defined.