Weak and Strong Integration of vector-valued functions - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-18T07:32:20Z http://mathoverflow.net/feeds/question/47721 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/47721/weak-and-strong-integration-of-vector-valued-functions Weak and Strong Integration of vector-valued functions Hadi 2010-11-29T21:13:09Z 2010-12-07T06:38:36Z <p>This is probably an elementary question, but outside my area of expertise, and I was unable to find any suitable reference:</p> <p>Suppose $f:X\to E$ is a continuous function from a compact spaces (endowed with a probability Borel measure $\mu$) to a locally convex, Hausdorff, complete topological vector space. We want to define a reasonable integral, i.e., $\int_Xfd\mu$. The question is : when exactly is that one needs weak integrals (e.g., the <em>Gelfand-Pettis</em> integral)? There is a strong integral called the <em>Bochner integral</em> which is typically defined for $E$ a Banach space; but it seems to me that its definition works at least for Frechet spaces. For a general locally convex, Hausdorff, complete TVS, when exactly does Bochner's approach, i.e., to define a strong integral by approximation of $f$ with simple functions (i.e., finite-valued measurable functions) fail?</p> http://mathoverflow.net/questions/47721/weak-and-strong-integration-of-vector-valued-functions/48327#48327 Answer by Johannes Hahn for Weak and Strong Integration of vector-valued functions Johannes Hahn 2010-12-04T23:25:55Z 2010-12-04T23:25:55Z <p>Hi.</p> <p>One issue with Bochner integration is that it does not include Riemann-integration. There are Banach-space-valued R-integrable functions that are not B-integrable (example: Consider $X:=\mathcal{l}^p([0,1])$ with $2\leq p &lt; \infty$ and $f:[0,1]\to X, f(t):=e_t$ where $e_t$ is the tupel with exactly one equal to 1 and all other components equal to 0.). The Gelfand-Pettis-Integral on the other hand includes both the Bochner- and the Riemann-Integral.</p> <p>One problem with B-integration is that you need functions that are almost separable valued (meaning: there is a nullset whose complement has separable image) in order to approximate them with simple functions and this may be a strong restriction. Another issue is that for certain applications the weak topologies just behave better than the strong ones so that Pettis-integration is the natural notion of integration in this cases.</p> http://mathoverflow.net/questions/47721/weak-and-strong-integration-of-vector-valued-functions/48360#48360 Answer by anton for Weak and Strong Integration of vector-valued functions anton 2010-12-05T11:44:19Z 2010-12-05T11:44:19Z <p>In the setting you name, the Bochner approach never fails. As the image $f(X)$ is compact, you construct a net of simple functions as follows. For index set, you take the set of all continuous seminorms $p$ on $E$. For each $p$, let $U_p$ be the open unit-ball defined by $p$. For $n\ge 1$ large enough, you cover $f(X)$ by finitely many translates on $\frac1n U_p$. Make the covering disjoint and make the vectors you translated by the values of a simple function $s_p$ which satisfies $\int_Xp(f-s_p)\,d\mu&lt;1$. The $(s_p)$ form a Cauchy-net, the limit of which is your desired integral.</p> http://mathoverflow.net/questions/47721/weak-and-strong-integration-of-vector-valued-functions/48540#48540 Answer by BR for Weak and Strong Integration of vector-valued functions BR 2010-12-07T06:38:36Z 2010-12-07T06:38:36Z <p>Anton's answer prompted me to pull out Rudin's Functional Analysis. Exercise 23 of Chapter 3 is to exhibit the Gelfand-Pettis integral of a continuous function with values in a Frechet space as a strong limit of "Riemann sums" (looks slightly stronger than the Bochner integral, in fact). I couldn't make sense of how to use the hint, so I decided to see what I could come up with. It's been a long time since I've done this sort of thing, so I may have done something stupid. </p> <p>We're going to assume $X$ is compact, $\mu$ is a probability Borel measure, $f$ is continuous, $E$ is locally convex and satisfies the technical criterion for the existence of Gelfand-Pettis integrals (closure of the convex hull of the image is compact). Then $f(X)$ is compact. If we also assume either $E$ is a metric space or $X$ is separable, then $f(X)$ is separable. Take a countable dense subset $e_i$ of $f(X)$. For a finite subset $B$ of seminorms and $\epsilon>0$, let $V(B,\epsilon,e_i)=\{e:p(e-e_i)&lt;\epsilon\}$. These form a basis for the topology of $E$ (and they are convex and balanced). </p> <p>First, we're going to construct a sequence of simple functions that converges to $f$, mimicking the example from measure theory.</p> <p>Since $f(X)$ is compact, we can form a cover with finitely many of the $V(B,\epsilon,e_i)$, call them $A_i$. Let $E_i=A_i-\bigcup_{j=1}^{i-1}A_j$. Set $X_i=f^{-1}(E_i)$, so the $X_i$ are a disjoint partition of $X$. </p> <p>Set $g_{B,\epsilon}(x)=\sum_i 1_{X_i}(x)e_i$ (where $i$ ranges over a finite set). Consider $f(x)-g_{B,\epsilon}(x)$. If $x\in X_i$, $f(x)\in E_i$, so for all $p\in B$, $p\big(f(x)-g_{B,\epsilon}(x)\big)=p\big(f(x)-e_i\big)&lt;\epsilon$. Convergence is easy from this.</p> <p>Now we will show that $\int_X g_{B,\epsilon}\ d\mu=\sum_i \mu(X_i)e_i$ converges. We "cheat" by showing it converges to the Gelfand-Pettis integral of $f$ (instead of e.g. showing that it is Cauchy). So consider $$\int_X f\ d\mu- \int_X g_{B,\epsilon}\ d\mu=\int_X f- g_{B,\epsilon}\ d\mu=\sum_i\int_{E_i}f-e_i\ d\mu$$</p> <p>On $E_i$, $f-e_i$ lies in $V(B,\epsilon)$, so $\int_{E_i}f-e_i\ d\mu$ is in the closure of $\mu(E_i)\cdot V(B,\epsilon)\subset \overline{V(B,\epsilon)}$ (since $V(B,\epsilon)$ is balanced) (this uses the estimate that the Gelfand-Pettis integral lies in the closure of the convex hull of the image, times the measure of the space). So $\sum_i\int_{E_i}f-e_i\ d\mu\subset \sum_i \overline{V(B,\epsilon)}\subset \overline{\sum_i V(B,\epsilon)}$.</p> <p>By some basic tvs theorems, since $V(B,\epsilon)$ forms a basis for the topology, we can choose $V(B,\epsilon)$ so that $\overline{\sum_i V(B,\epsilon)}$ is inside any given neighborhood of $0$ (see p.10-11 of Rudin).</p> <p>So Bochner integrals of continuous functions exist whenever Gelfand-Pettis integrals exist (plus a separability assumption), unless I made a mistake. Even though I left it in, I don't think that I used separability in an essential way (I could have just chosen an uncountable dense subset $e_i$ and proceeded). That requirement for strong measurability (almost-separably valued) may be an artifact of using sequences instead of nets. </p>