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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

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

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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.

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    $\begingroup$ 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 $K$? $\endgroup$ – ougoah Dec 8 '17 at 7:50
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    $\begingroup$ 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 '17 at 8:22
  • $\begingroup$ That's a good point, thanks. So it really is a red herring... $\endgroup$ – ougoah Dec 8 '17 at 14:10
  • $\begingroup$ Thanks, I missed the assumption that $h \in K$. I modified my example following @gerw's suggestion. $\endgroup$ – Nate Eldredge Dec 8 '17 at 14:40

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