I am trying to understand the proof of the Fundamental Theorem of Calculus for Banach space-valued functions, and in particular, how the Theorem of Hahn-Banach is being applied there.

In the following, let $f:[x,y]\to E$ be a $C^1$-function taking values in a Banach space $E$, where $[x,y]\subset\mathbb R$ with $y\gt x$. Here, I consider $\mathbb R$ together with Lebesgue measure $\mu$.

Now the aim is to show that, $$\int_x^yf'\,d\mu=f(y)-f(x) ,$$ where the integral on the left is that of Bochner. In the proofs I have seen so far, we proceed as follows: for any $\phi\in E^*$ we have that,

$$\phi\left(\int_x^yf'd\mu\right)=\int_x^y\phi\circ f'd\mu=\phi\left(f(y)-f(x)\right).$$

In the first equality we have used the theorem which tells us that any bounded linear operator commutes with the Bochner integral (this is very similar to the more familiar Theorem of Hille), and in the second we have used the ordinary Fundamental Theorem of Calculus.

In the proofs I have looked at, we are then meant to apply the Hahn-Banach Theorem. It seems to me that we are meant to apply this Corollary:

Let $X$ be a normed vector space. If $x\in X$ is nonzero then there exists a $\psi\in X^*$ such that $\|\psi\|=1$ and $\psi(x)=\|x\|$.

Since the above chain of equalities holds for all $\phi\in E^*$ it will hold, in particular, for that functional provided by the above Corollary. For this functional, it then follows that,


but this is where I get a little unsure on the rationale behind applying Hahn-Banach. First of all, is this the right version of the Hahn-Banach Theorem we are meant to apply? Secondly, what, exactly, am I using it to show or deduce - that is to ask, how does its application get me back to the derivation of the original aim?

  • $\begingroup$ I don't really see the equality where you used the ordinary fundamental theorem of calculus. Why is a antiderivative of $\phi \circ f^{'}$ given by $\phi\circ f$? $\endgroup$
    – user501184
    Apr 28, 2019 at 11:45
  • $\begingroup$ $\phi: E \to \mathbb{K}$ is linear. It's differential (for a map to the field $\mathbb{K}$, the "gradient"), is itself. Note that in one variable $(f\circ g)' = g' \cdot f' \circ g$ but in several variables and a fortiori in a Banach space $d(f\circ g)= df \circ dg$. Here $(\phi\circ f)'=\phi'\circ f' = \phi \circ f'$ $\endgroup$
    – Noix07
    Dec 20, 2019 at 12:39

1 Answer 1


We have $\phi (\int_x^{y}f'd\mu -(f(y)-f(x))=0$ for all $\phi \in E^{*}$. So the only thing left is to show that if $u \neq 0$ in $E$ then there exists $\phi \in E^{*}$ such that $\phi (u) \neq 0$. This is where you can apply the Hah n Banach argument: there exists $\phi \in E^{*}$ such that $\|\phi\|=1$ and $\phi (u) =\|u\| >0$ if $u \neq 0$.

  • $\begingroup$ We want to show that $\int_x^{y}f'd\mu -(f(y)-f(x))=0$. Is the idea to use this Corollary of Hahn-Banach to obtain a contradiction? Namely that for $\int_x^{y}f'd\mu -(f(y)-f(x))\in E$ there does not exist $\phi\in E^*$ such that $\|\phi\|=1$ or $\phi(\int_x^{y}f'd\mu -(f(y)-f(x)))\neq0$? $\endgroup$ Apr 8, 2019 at 10:27
  • 1
    $\begingroup$ Yes. I have taken $u$ to be $\int_x^{y}f'd\mu-(f(y)-f(x))$. To prove that this is $0$ (by contradiction) assume $u \neq 0$. Then you get $\phi$ as mentioned you. But we already know that $\phi (u)=0$ so we cannot have $\phi (u)=\|u\|$. $\endgroup$ Apr 8, 2019 at 10:31

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