I am reading a lecture note about Radon Riesz theorem, the resources is here: https://faculty.etsu.edu/gardnerr/5210/notes/Radon-Riesz.pdf
At page 4, fourth line from the bottom, it says convergence in measure. But if this result is implied by weak convergence of $f_n$. I can’t proof it.
I have tried some methods, but this lead to a question about the definition of weak convergence. In the definition, which of the following one is exactly the real definition of weak convergence?
- for every $T\in X^*$, for all $\varepsilon>0$ May related to $T$, there is an $N$, such that for all $n>N$, $d(Tf_n, Tf)<\varepsilon$.
- for all $\varepsilon>0$, there is an $N$, such that for all $n>N$ and all $T$, $d(Tf_n, Tf)<\varepsilon$.
I checked some book but they generally says like
$$Tf_n\to Tf,\qquad \text{for all } T\in X^*$$
This makes curious for me.
Finally I almost forgot, if the definition is the second one I have proved: weak convergence implies convergence in measure, but if the definition is the first one, the prove failed, I also need a prove or a counterexample that weak convergence can’t imply convergence in measure, I also want to know how the lecture note get the result about convergence in measure if there is really an counterexample. THX.
Best Answer
The second one is false. To see this, if $T$ is a (fixed) continuous linear functional, then any constant multiplied by $T$ is also a continuous linear functional. Then the second statement is essentially asking for $$|cT(f_n) - cT(f)| = c|T(f_n) - T(f)| \leq \epsilon$$ for any constant $c$.
The first one is (almost) correct, the $\epsilon$ is still arbitrary, just now $N$ is allowed to depend on $T$.
In general, weak convergence does not imply convergence in measure, take $f_n(x) = \chi_{[n, n+1)}(x)$ which weakly converges to zero in $L^p(\mathbb{R})$ for $1<p<\infty$, but fails to converge in measure .