By the fundamental theorem of calculus I mean the following.

Theorem: Let $B$ be a Banach space and $f : [a, b] \to B$ be a continuously differentiable function (this means that we can write $f(x + h) = f(x) + h f'(x) + o(|h|)$ for some continuous function $f' : [a, b] \to B$). Then

$$\int_a^b f'(t) \, dt = f(b) - f(a).$$

(This integral can be defined in any reasonable way, e.g. one can use the Bochner integral or a Riemann sum.)

This theorem can be proven from Hahn-Banach, which allows you to reduce to the case $B = \mathbb{R}$. However, Hahn-Banach is independent of ZF.

Recently I tried to prove this theorem without Hahn-Banach and found that I couldn't do it. The standard proof in the case $B = \mathbb{R}$ relies on the mean value theorem, which is not applicable here. I can only prove it (I think) under stronger hypotheses, e.g. $f'$ continuously differentiable or Lipschitz.

So I am curious whether this theorem is even true in the absence of Hahn-Banach. It is likely that I am just missing some nice argument involving uniform continuity, but if I'm not, that would be good to know.

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    $\begingroup$ The motivation, in case anyone is wondering, is that I am still trying to prove standard facts about Banach space-valued holomorphic functions without Hahn-Banach, and I think this is the only nontrivial technical step. Everything else should more or less follow from adaptations of the usual proofs. $\endgroup$ Aug 22, 2012 at 3:33

3 Answers 3


I believe that one of the standard proofs works.

  1. Let $F(x) := \intop_a^x f^\prime (t) dt$. Then $F$ is differentiable and its derivative is $f^\prime$ due to a standard estimate that has nothing to do with AC.

  2. $(F-f)^\prime = 0$, hence it is constant. This boils down to the one-dimensional case: just consider $g := \Vert F-f-F(a)+f(a) \Vert$. It is a real-valued function with zero derivative, and $g(a)=0$, so we can use the usual "one-dimensional" mean value theorem.

  • $\begingroup$ Regarding 1, I also tried and failed to prove that $F$ is differentiable (again the standard proof that I am aware of uses the mean value theorem). Please elaborate if you actually have a complete proof of this. $\endgroup$ Aug 22, 2012 at 4:08
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    $\begingroup$ Well, $\Delta F(x) = f^\prime(x) \Delta x + \intop_x^{x+\Delta x} (f^\prime(t) - f^\prime(x)) dt$. You say that you can define the integral using Riemann sums: Ok, then every Riemann sum is bounded in norm by $\Delta x \cdot \sup_{[x, x+\Delta x]} \Vert f^\prime(t) - f^\prime(x) \Vert$. $\endgroup$ Aug 22, 2012 at 4:19
  • $\begingroup$ How do you show $g$ is differentiable? Perhaps one should look at its projection under linear functionals. $\endgroup$
    – Miheer
    Jun 20, 2019 at 20:49

Claim: Let $g:[a,b]\to B$ be differentiable, with $g'(t)=0$ for all $t\in[a,b]$. Then $g(t)$ is a constant.

Proof: Fix $\epsilon>0$. For each $t\in[a,b]$, we can find $\delta_t>0$ such that $0<|h|<\delta_t\Rightarrow\|g(t+h)-g(t)\|<\epsilon|h|$. The open intervals $(t-\delta_t,t+\delta_t)$ cover $[a,b]$, so there is a finite subcover $\{(t_i-\delta_{t_i},t_i+\delta_{t_i}):1\leq i\leq N\}$. We may choose our labeling so that $t_1<t_2<\ldots<t_N$. Now, we should be able to find points $x_0,\ldots,x_N$ with $x_0=a<t_1<x_1<t_2<\ldots<t_N<x_N=b$, satisfying $|x_i-t_i|<\delta_i$ and $|x_{i-1}-t_i|<\delta_i$ for $1\leq i\leq N$. Now, \begin{eqnarray*} \|g(b)-g(a)\|&=&\left\|\sum_{i=1}^N(g(x_i)-g(t_{i}))-(g(t_i)-g(x_{i-1}))\right\|\\ &<&\sum_{i=1}^N \epsilon (x_i-t_i)+\epsilon(t_i-x_{i-1})\\ &=&\epsilon(b-a) \end{eqnarray*} Since $\epsilon$ is arbitrary, we have $g(b)=g(a)$. This argument works on any subinterval, so $g(t)$ is constant.

We can apply the above with $g(x):=\int_a^xf'(t)\,dt-(f(x)-f(a))$. I believe it can be checked directly that $g'(x)=0$ for all $x$, and that $g(a)=0$, so that $g(x)$ must be identically 0.

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    $\begingroup$ Okay, I think I believe your claim, but I couldn't prove that $\int_a^x f'(t) \, dt$ is differentiable either. Can you prove this? $\endgroup$ Aug 22, 2012 at 4:10

It seems that all suggestions above work under the assumption that $f$ is continuous and piece wise continuously differentiable in order to be able to use standard uniform continuity type arguments. Here is how it could be done under the assumption above:

It is enough to assume that $f'$ is also continuous, The strong measurability of $f'$ is straight forward and so is the integrability of $\|f'\|$. This in turn implies that $f'$ is integrable. The sequence $$f'_n(t)=\sum^{(2^n-1)\wedge b}_{k=(-2^n)\vee a}f'\big(\frac{k}{2^n}\big)\mathbb{1}_{\big(\frac{k}{2^n},\frac{k+1}{2^n}\big]}(t)$$ For instance can be use to approximate $f'$ over $[a,b]$ uniformly. The slightly tricky part is to show $f(b)-f(a)=\int^b_a f'$. For this, We observe that $$\frac{1}{h}\Big\|\int^{t+h}_{t}\big(f'(s)-f'(t)\big)\,ds\Big\|\leq \sup_{x,y\in[t,t+h]}\|f'(x)-f'(y)\|$$ Which can be made small for sufficiently small $h$ by the uniform continuity of $f'$ over $[a,b]$. Similar argument can be used for $[t-h,t]$. ALl this shows that $F:t\mapsto\int^t_af'(s)\,ds$ is differentiable and that $F'(t)=f'(t)$. The conclusion then follows from the mean value theorem for Banach spaces: $$\|F(b)-(f(b)-f(a))-F(a)\|\leq \sup_{a\leq x\leq b}\|F'(x)-f'(x)\||b-a|=0$$ More general situations, for instance, when all one assumes is that $f$ is absolutely continuous over $[a,b]$ and then get as consequence that

  • $f$ is differentiable almost everywhere,
  • that $f'$ is integrable (a la Bochner),
  • and that $f(b)-f(a)=\int^a_b f(t)\,dt$

would require additional assumptions on the Banach space $B$. For instance, that its dual $B^*$ is separable. The use of Hanh-Banach's theorem then would seem to be inevitable.


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