I'm studying for an analysis written exam and came across this problem:

Suppose that $\{f_k\}$ and $\{g_k\}$ are two sequences of functions in $L^2([0, 1])$. Suppose that $$\|f_k\|_2 \leq 1\ \text{for all}\ k$$ where the $L^p$ norm $\|\cdot\|_p,\ 1 \leq p \leq \infty$, is defined with respect to the standard Lebesgue measure $\mu$ on $[0, 1]$. Suppose further that there exist $f, g \in L^2([0, 1])$ such that $f_k(x) \to f(x)$ for a.e. $x \in [0, 1]$ and that $g_k \to g$ in $L^2([0, 1])$.

Prove that $f_kg_k \to fg$ in $L^1([0, 1])$.

I understand what the problem is asking and am familiar with all these types of convergence, but I can't figure out where to start. I would greatly appreciate any hints, ideas, or suggestions of what theorems would be useful!

  • First, why is it the case that $fg\in L^1[0,1]$? (Hint: Cauchy—Schwarz).

  • Then, note that $$\lvert f_kg_k - fg\rvert = \lvert f_kg_k - f_kg + f_kg - fg\rvert \leq \lvert f_kg_k - f_kg \rvert + \lvert f_kg - fg\rvert \tag{1} $$ and then that, by Cauchy—Schwarz, $$\begin{align} \int_{[0,1]}\lvert f_kg_k - f_kg \rvert &= \int_{[0,1]}\lvert f_k\rvert \lvert g_k - g \rvert \leq \sqrt{ \int_{[0,1]} f_k^2 } \cdot \sqrt{ \int_{[0,1]} (g_k - g)^2} \\ &\leq 1 \cdot \lVert g_k - g\rVert_2\tag{2} \end{align}$$ (can you justify all inequalities?)

    Similarly, $$\begin{align} \int_{[0,1]} \lvert f_kg - fg\rvert&=\int_{[0,1]} \lvert f_k - f\rvert\lvert g\rvert \leq \sqrt{ \int_{[0,1]} g^2 } \cdot \sqrt{ \int_{[0,1]} (f_k - f)^2}\\ &\leq \lVert g\rVert_2 \cdot \lVert f_k - f\rVert_2 \tag{3} \end{align}$$

    Can you now (i) conclude from (1), (2), and (3)? (ii) justify the few parts not entirely spelled out? and importantly, (iii) see where the assumption on the $f_k$'s being uniformly bounded was used, and what may fail in the proof without it?

  • $\begingroup$ Ah, I see! Yes, that makes total sense. Thank you! $\endgroup$ – CFish Jun 20 '17 at 0:07
  • $\begingroup$ @CFish You're welcome! $\endgroup$ – Clement C. Jun 20 '17 at 1:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.