Suppose that a sequence of bounded and continuous functions $f_n$ converges uniformly to $f_1$ and $f_n$ converges to $f_2$ in $L^2$ sense, then how to show $f_1= f_2$ a.e.?

I tried the following: let $A_\epsilon = \{x:|f_1(x)-f_2(x)|>\epsilon\}$, then $m(A_\epsilon) < m(|f_n - f_1|>\epsilon) + m(|f_n - f_2|>\epsilon)$. Let $n$ go to infinity, then the first part of RHS goes to zero by uniform convergence, but I cannot do anything to $L^2$-convergence.

Can anyone show me how to solve this question? Thanks in advance .

  • 2
    $\begingroup$ You are on the right track: use $m(|f_n-f_2|\gt\epsilon)\leqslant\epsilon^{-2}\|f_n-f\|_2^2$. $\endgroup$ – Did Aug 25 '12 at 15:17
  • $\begingroup$ @Norbert Asked 5 hours ago is a bit soon for a question to be declared unanswered, don't you think? $\endgroup$ – Did Aug 25 '12 at 20:10
  • $\begingroup$ I think people shy to post answer that you have already gave in first comment. I really don't like common practice of posting answers as comments $\endgroup$ – Norbert Aug 25 '12 at 22:24
  • $\begingroup$ @Norbert What you say does not correspond to my experience, as I have seen countless examples of the opposite happening on this site. Anyway, your second comment forces me to interpret your first one quite differently than I first did, and in a way which I really don't like. $\endgroup$ – Did Aug 25 '12 at 22:48
  • $\begingroup$ @did So what you don't like? $\endgroup$ – Norbert Aug 25 '12 at 23:49

Markov's inequality does the job: we get that for each $\varepsilon>0$, $$\lambda\{x,|f_n(x)-f_2(x)|>\delta\}\leqslant \frac 1{\varepsilon^2}\lVert f_n-f_2\rVert_{L^2}^2,$$ hence following the notations in the OP, we get $$\lambda(A_{2\varepsilon}\cap [-N,N])\leqslant \lambda(\{x,|f_n(x)-f_1(x)|>\varepsilon\}\cap [-N,N])+\frac 1{\varepsilon^2}\lVert f_n-f_2\rVert_{L^2}^2,$$ hence $$\lambda(A_{2\varepsilon}\cap [-N,N])\leqslant 2N\cdot \left[\sup_{[-N,N]}|f_n-f_1|>\varepsilon\right]+\frac 1{\varepsilon^2}\lVert f_n-f_2\rVert_{L^2}^2,$$ where $[P]$ is when when $P$ is satisfied and $0$ otherwise.

When $N$ and $\varepsilon$ are fixed, the RHS goes to $0$ as $n$ goes to infinity. Hence, $\lambda(A_{2\varepsilon}\cap [-N,N])=0$ for all $N$ and $\varepsilon$, giving the wanted conclusion.


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.