I have these in books without proof, mostly as a corollary. I was wondering if I could get a proof.

Suppose $$\lim_{n\to \infty} \int_0^1 f_ng dx = \int_0^1 fg dx$$ for all $g\in L^2(0,1)$, where $f_n, f \in L^2(0,1)$. Then there exists a constant $K$ such that $\|f_n\|_{L^2} \leq K \lt \infty$ for all $n$.


Banach-Steinhaus tells us that a family of linear functionals on a Banach space is either unbounded on a dense $G_\delta$ set or is uniformly bounded in norm. Since this family converges weakly we have for any $g \in L^2$ that $\sup_n \langle f_n , g \rangle < \infty$ and therefore the family of functionals is uniformly bounded in norm.

  • $\begingroup$ Is it possible to elaborate a bit? $\endgroup$ – Josh May 6 '12 at 13:04
  • $\begingroup$ @Josh on which part? $\endgroup$ – Chris Janjigian May 6 '12 at 15:50
  • $\begingroup$ The conclusion. $\endgroup$ – Josh May 6 '12 at 17:08
  • 1
    $\begingroup$ @Josh Banach-Steinhaus (aka the uniform boundedness principle) says that exactly one of two things is true: either we can find a lot of functions (in particular, one exists) in $L^2$ for which $\sup_n \langle f_n , g\rangle = \infty$ or $\|f_n\| \leq C$ for some uniform constant $C$. But the $f_n$ sequence converges weakly, so that for any $g$ $\lim_{n \to \infty} \langle f_n , g \rangle$ exists. A convergent sequence of reals is bounded and so in particular we know that the set of $g$ for which $\sup_n \langle f_n , g \rangle = \infty$ is empty, so we must have a uniform norm bound. $\endgroup$ – Chris Janjigian May 6 '12 at 21:06
  • $\begingroup$ What does the symbol $\langle f_n , g\rangle$ mean? $\endgroup$ – Josh May 7 '12 at 21:35

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.