Analysis Problem of the Day 37

Today’s problem appeared as Problem 1 on the UCLA Spring 2011 Analysis Qual:

Problem 1: Define what it means to say that $f_n$ converges to $f$ weakly in $L^2([0,1]).$ Define the primitive functions $F_n(x) = \int_0^x f_n(t)dt, F(x) = \int_0^x f(t)dt.$ Show that $F_n, F \in C([0,1])$ and $F_n \to F$ in $C([0,1]).$

Solution: To say $f_n \rightharpoonup f$ in $L^2([0,1])$ (which is a Hilbert space) is to say that for every $g \in L^2([0,1]),$ $\int_0^1 f_n(t) \overline{g(t)}dt \to \int_0^1 f(t)\overline{g(t)}dt.$ Notice that $L^2([0,1]) \subset L^1([0,1]),$ so $f_n, f$ are integrable. Thus, the primitive functions are well-defined and continuous, since by absolute continuity of integrals, for all $\epsilon>0,$ there exists $\delta>0$ such that $\int_K |f| < \epsilon$ whenever $\mu(K)<\delta$ for $f \in L^1.$ It follows that $F_n,F$ are in fact absolutely continuous. Note that since $f_n \rightharpoonup f,$ it follows that $F_n \to F$ pointwise. To show uniform convergence, we note that $\sup_n \|F_n\|_\infty \leq \sup_n \|f_n\|_1 \leq \sup_n \|f_n\|_2 < \infty$ and $|F_n(x)-F_n(y)| = \left|\int_x^y f_n(t) dt\right| \leq (\sup_n \|f_n\|_2) |x-y|^{\frac12},$ showing that $\{F_n\}$ is a uniformly bounded equicontinuous family of functions. By Arzela-Ascoli, it follows that every subsequence of $\{F_n\}$ has a further subsequence that converges uniformly to $F,$ i.e. $F_n$ converges uniformly to $F.$

Remark: In general, a sequence of absolutely continuous functions converging uniformly do not have to converge to an absolutely continuous function, since $C^\infty([0,1])$ is dense in $C([0,1])$ by Stone-Weierstrass. It is true if we impose some convergence on the derivative, however, such as convergence in the Sobolev space $W^{1,1}.$

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