Let $(H,||\cdot||_H)$ be a Banach space and $K$ a (not necessarily closed) subspace. Suppose that $K$ is a Banach space under another norm $||\cdot||_K$, which satisfies

$$||x||_H\leq ||x||_K$$

for all $x\in K$. Let $(S,\mu)$ be a measure space and $f:S\rightarrow K$ a strongly measurable function in the sense of Bochner, with respect to both $||\cdot||_K$ and $||\cdot||_H$. That is, there exist sequences of simple functions $\phi_n$ and $\psi_n$ on $S$ taking values in $K$ and $H$ respectively such that

$$\lim_{n\rightarrow\infty} ||f(s) - \phi_n(s)||_{K} = \lim_{n\rightarrow\infty} ||f(s) - \psi_n(s)||_{H} = 0$$

for $\mu$-almost all $s\in S$.

Suppose now that $f$ satisfies

$$\int_S ||f(s)||_H\,d\mu(s) < \infty,$$

so that, by a criterion of Bochner, $f$ is integrable as a function with values in $H$. Denote $$h := \int_S f(s) d\mu(s).$$

Question: If $h\in K$, then does it follow that $f$ is Bochner-integrable in $K$? That is, does there exist a sequence of simple functions $\phi_n$ such that

$$\lim_{n\rightarrow\infty}\int_S ||f(s) - \phi_n(s)||_{K}\,d\mu(s) = 0?$$


1 Answer 1


The Bochner integral is a red herring here. Let $S = \mathbb{N}$ and let $\mu$ be counting measure. Then a "strongly measurable function" is just a sequence, and it is "Bochner integrable" iff it is absolutely summable (in $H,K$ respectively).

If the $H$ and $K$ norms are not equivalent (which is necessarily the case when $K$ is not closed) then we can find a sequence $y_n \in K$ with $\|y_n\|_H \le 2^{-n}$ but $\|y_n\|_K \ge 2^n$. Set $x_1 = y_1$, $x_{n+1} = y_{n+1} - y_n$, and think of the function $f(n) = x_n$. Note that $\|x_n\|_H \le 2^{-(n-1)}$ and $\|x_n\|_K \ge 2^{n-1}$.

Now $x_n$ is absolutely summable in $H$-norm and $\sum_n x_k = 0$ because the sum telescopes. That is to say that $f$ is Bochner integrable in $H$ and the value of the integral $\int_S f\,d\mu$ is 0 which is in $K$.

On the other hand, $\sum_n x_n$ diverges in $K$, which is to say that the Bochner integral in $K$ does not exist.

  • 1
    $\begingroup$ This shows that there are functions with values in $K$ that are Bochner-integrable in $H$ but not in $K$. But the question was, can this happen if the integral evaluates to an element of $K$? $\endgroup$
    – geometricK
    Dec 8, 2017 at 7:50
  • 3
    $\begingroup$ But you can modify Nate's idea to get a series $\sum_{n=1}^\infty x_n$ which converges absolutely in $H$ to $0$, but it diverges in $K$. In the same spirit: if $K \ni y_n \to 0$ in $H$, you still might have the divergence of $y_n$ in $K$ although the limit $0$ belongs to $K$. $\endgroup$
    – gerw
    Dec 8, 2017 at 8:22
  • $\begingroup$ That's a good point, thanks. So it really is a red herring... $\endgroup$
    – geometricK
    Dec 8, 2017 at 14:10
  • $\begingroup$ Thanks, I missed the assumption that $h \in K$. I modified my example following @gerw's suggestion. $\endgroup$ Dec 8, 2017 at 14:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.