Am I correct with usage of this generalised Dominated Convergence lemma: a sequence $(f_n)$ in $L^1$ on a bounded domain is strongly convergent if and only if $(f_n)$ is uniformly integrable and $(f_n)$ converges in measure, or pointwise almost everywhere?

By Dunford Pettis I know that $(f_n)$ is uniformly integrable if and only if $(f_n)$ is weakly convergent in $L^1$.

My question is, is it possible to deduce $L^p$ convergence of a bounded sequence $(f_n)\subset L^p$, by showing that, for all $\phi\in L^\infty$: $$1. \int |f_n |^p(x) \phi (x) dx\to \int |f |^p(x) \phi (x) dx$$ and $$2. \,\, f_n\to f \text{ in measure} $$ which would imply the uniform integrability of $|f_n-f|^p$, and hence the desired convergence? I would attempt to use the Vitali lemma, so that on a bounded domain $\Omega$, $(|f_n|^p)$ uniformly bounded in $L^1(\Omega)$, existence (and finiteness) of the pointwise a.e limit and uniform integrability of $|f_n-f|^p$ allow passage to the limit $$\lim_n \int_\Omega |f_n-f|^p dx = \int_\Omega\lim_n |f_n-f|^p dx =0.$$

Furthermore, how limited would an approximation argument for 1., with $\phi\in C_c^\infty$? Is there a counterexample with a uniformly integrable sequence, pointwise converging, that does not converge in $L^p$? I ask this second question because of the equivalence of distributional convergence to weak convergence in $L^p$ if the sequence (and limit) are already bounded in $L^p$.

Note: I am using the following definition of UI: $\forall\varepsilon>0, \exists g$ an integrable function such that $$\sup_{n}\int_\Omega(|f_n|-g)^+\mathrm dx\lt\varepsilon.$$

  • $\begingroup$ I don't know qnd I really have no time to think at this now, but I'd like to leave a commentary anyway. Weak convergence in $L^1$ space and in $L^p$ space are very different things. The reason is that $L^1$ space embeds in the space of measures, so the "right" setting for weak convergence of $L^1$ functions really is seeing them as measures. There is nothing analogous for $L^p$. (For instance, note that you cannot "converge to the delta distribution" in $L^p$ space. The scaling is just wrong.) HTH even if it is rather vague; if you find it confusing I'll erase this comment $\endgroup$ – Giuseppe Negro May 27 '14 at 13:56
  • $\begingroup$ Thanks, I understand. I was trying to think of a way to prove a higher uniform integrability than $L^1$, but cannot find anything equivalent to the definition. $\endgroup$ – michek May 27 '14 at 13:59
  • $\begingroup$ I am trying to make use of the equivalence that for a uniformly bounded sequence $(f_n)$ in $L^p, 1<p<\infty$, with $f\in L^p$, then $f_n\to f$ weakly in $L^p$ iff $f_n\to f$ weakly-* in the sense of measures (or even $f_n\to f$ in $\mathcal{D}'$). $\endgroup$ – michek May 27 '14 at 14:23
  • $\begingroup$ Is your measure space finite? Otherwise, what definition of uniform integrability do you use? Because with the definition I know, the sequence $f_n = \chi_{[n,n+1]}$ on $\Bbb{R}$ with Lebesgue measure should be a counter example to your initial statement. $\endgroup$ – PhoemueX May 27 '14 at 16:20
  • $\begingroup$ Yes it is finite. I will edit to include the definition. $\endgroup$ – michek May 27 '14 at 16:25

I just noted that the statement you want to prove is actually trivial, because if

$$\int |f_n - f|^p \cdot \phi \, dx \rightarrow 0$$

for all $\phi \in L^\infty$, you can use this with $\phi \equiv 1$ which is then nothing but convergence in $L^p$.

EDIT: If you instead require $\phi \in C_c^\infty(\Omega)$, the statement becomes false in general. As a counterexample consider $p=2$, $\Omega = (0,1)$ and $f_n = \sqrt{n} \cdot \chi_{(0,1/n)}$. Of course, in this case, the sequence is not uniformly integrable.

  • $\begingroup$ Then maybe I am misunderstanding Dunford Pettis: what should it mean to be relatively compact in the weak $L^1$ topology? Wouldn't this trivially imply strong $L^1$ convergence as you have shown? $\endgroup$ – michek May 28 '14 at 9:56
  • $\begingroup$ The point is that weak $L^1$ convergence means $\int (f_n - f) \cdot \phi dx \rightarrow 0$ for all $\phi \in L^\infty$. Note the missing absolute value around $f_n - f$. You could choose the $\phi$ so that the absolute value does not matter, but this requires $\phi = \phi_n$ in general, so that the statement with the absolute values is strictly stronger. $\endgroup$ – PhoemueX May 28 '14 at 10:26
  • $\begingroup$ I think I see the problem now. Could one argue this holds for the positive and negative parts of $f_n-f$? $\endgroup$ – michek May 28 '14 at 10:29
  • $\begingroup$ Perhaps weak convergence of the $p-th$ powers, plus convergence of their norms, would give the $L^p$ convergence for all reflexive $p's$. $\endgroup$ – michek May 28 '14 at 10:37
  • $\begingroup$ Do you mean this? If $f_n \rightarrow f$ weakly in $L^p$ and $\Vert f_n \Vert_p \rightarrow \Vert f \Vert_p$ (and $1<p<\infty$), then already $f_n \rightarrow f$ strongly in $L^p$. If so, have a look at math.stackexchange.com/questions/163209/… . $\endgroup$ – PhoemueX May 28 '14 at 11:42

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