# Family of Pettis integrals functions “uniformly approximated” by sums

In this book (proof of $$4.1.3.$$ Lemma. exactly), one can find this passage, that I tried to rephrase here:

Let $$f:I\times E\rightarrow E$$ a Pettis integrable function, where $$I:=[0,T]\subset \mathbb{R}$$, and $$E$$ is a Banach space. Let $$\Omega$$ be a bounded, equicontinuous subset of $$\mathcal{C}(I,E)$$.

Suppose that $$f(.,y(.)),\;y\in \Omega$$ is equicontinuous.

Then, the integrals of these functions $$\int_{0}^{t}f(s,y(s))ds,\;y\in \Omega$$ can be uniformly approximated by integral sums $$\frac{t}{n} \sum_{i=1}^{n} f\left(s_{i}, y\left(s_{i}\right)\right), \quad s_{i}=i \frac{t}{n}, y \in \Omega$$

My first question is: what this "uniformly approximated" refers to?

Secondly, I am looking for a proof of this result, and it will be great if someone give me a reference to include in an article.

EDIT: As pointed by @Jochen Wengenroth in comments, this is not true in general. In the book they suppose that $$f(.,y(.)),\;y\in \Omega$$ is also equicontinuous, I forgot to mention that, and i'm sorry for that!

• I do not have access to the book but I doubt that this result is true: For $E=\mathbb R$ the Pettis integral is the Lebesgue integral and for $f(s,y)=1_{\mathbb R \setminus \mathbb Q}(s)$ the integral $\int_0^t f(s,y(s))ds$ is $t$ whereas the Riemann sums are $0$. The story is of course different for continuous $f$ -- although it appears to me that equicontinuity of the family $\Omega$ would be a more natural assumption in order to have the convergence uniform with respect to $y\in\Omega$. – Jochen Wengenroth Oct 6 at 8:10

As @Jochen commented, the result isn't true as originally stated. The book demands that the functions $$(s\mapsto f(s,y(s)) : y\in \Omega)$$ are equicontinuous, i.e. that for all $$\varepsilon>0$$, there exists $$\delta>0$$ such that for all $$y\in \Omega$$ and $$u,v\leq t$$, $$|u-v|<\delta \implies \lVert f(u,y(u))-f(v,y(v))\rVert<\varepsilon$$ ($$\delta$$ is not allowed to depend on $$y$$, or on $$u$$ or $$v$$).
"Uniformly approximated" here means that the error in the approximation tends to zero uniformly in $$y\in \Omega$$, i.e. $$\sup_{y\in \Omega} \Big\lVert\frac{t}{n} \sum f(s_i,y(s_i))- \int f(s,y(s))ds\Big\rVert\to 0.$$
To prove this, fix $$\varepsilon>0$$ and choose $$N$$ such that $$\lVert f(u,y(u))-f(v,y(v))\rVert \leq \varepsilon$$ whenever $$|u-v| ($$N$$ can be chosen independent of $$y$$ thanks to equicontinuity). Then for $$n\geq N$$, \begin{align*}\Big\lVert \frac{t}{n} \sum f(s_i,y(s_i))-\int f(s,y(s))ds\Big \rVert &= \Big\lVert \sum_{i=1}^n \int_{s_{i-1}}^{s_i} (f(s_i,y(s_i))-f(s,y(s)))ds\Big\rVert \\ &\leq \sum_{i=1}^n |s_i-s_{i-1}|\!\!\!\sup_{s\in [s_{i-1},s_i]} \!\!\lVert f(s_i,y(s_i))-f(s,y(s))\rVert \\ &\leq t\varepsilon,\end{align*} where we've used the triangle inequality and that the integral of the norm upper bounds the norm of the integral. We can take the supremum over $$y\in \Omega$$ because the right side doesn't depend on $$y$$, then as $$\varepsilon >0$$ was arbitrary this implies the result.
One thing I haven't checked which you should is whether what I've stated is exactly what the authors mean by equicontinuity. It shouldn't be far off, but the Pettis integral is usually associated with the weak topology, so it may be that the authors have some definition involving $$|\phi(f(u,y(u)))-\phi(f(v,y(v)))|$$ with $$\phi$$ in the dual space of $$E$$. If they do have something like that, more or less the same proof should work if you just apply $$\phi$$ everywhere.
• Oops I argued as if the range of $f$ was $\mathbb{R}$ rather than a general Banach space. The idea essentially works, you just need to take the norm of the difference and put the supremums outside. I can fix it later if needed – Kweku A Oct 6 at 9:16
• I will be grateful if you can give more details, it seems that you didn't use the fact that $\Omega$ is equicontinuous, did you? – Motaka Oct 6 at 9:41
• I've updated to correct the range of $f$ and also emphasise where equicontinuity is clear – Kweku A Oct 7 at 11:34