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In this question a user asks if pointwise convergence implies convergence in $L^p$. I would have thought that the answer is yes. I am not experienced with measure theory, which is how that question is framed. The following statement seems to assert that p.w. convergence implies convergence in $L^p$: $$ \lim_{n\to \infty} ||f_n - f||_{L^p(\Omega)}^p = \lim_{n\to \infty} \int_\Omega |f_n(x)-f(x)|^p dx = \int_\Omega |\lim_{n\to \infty} f_n(x)-f(x)|^p dx = \int_\Omega |0|^p dx = 0. $$ But the answers to the other post say that p.w. convergence does not imply convergence in $L^p$, so what am I missing?

ManUtdBloke
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    You can't always interchange integral with limit. – Angina Seng Sep 01 '17 at 06:44
  • So you are saying that even though the norm is a continuous function we can't say $$\lim_{n\to \infty} ||f_n|| = ||\lim_{n\to \infty} f_n||$$? I always thought this property characterizes continuous functions! – ManUtdBloke Sep 01 '17 at 07:23
  • Yes, the norm is always continuous. But this would require that you already know $f_n \to f$ in $L^p$ (and this is what you are trying to prove). – gerw Sep 03 '17 at 17:01

2 Answers2

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For $\Omega=\Bbb R$ and Lebesgue measure I like this example. $$f_n(x)=e^{-(x-n)^2}.$$ Then $f_n\to0$ pointwise, but $$\|f_n\|_p=\|f_1\|_p>0.$$

Angina Seng
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  • Ok that's a nice example. So I take it this example is one of the cases where we can't interchange the integral and the limit. So what is it about the sequence of functions in this case that prevent us from doing that? – ManUtdBloke Sep 01 '17 at 06:53
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    You should look up the monotone convergence theorem and the dominated convergence theorem. Both are highly related and will tell you additional conditions under which you will get $L^p$ convergence from pointwise convergence. It is quite insightful to see how those additional conditions are violated by some counterexamples. – mlk Sep 01 '17 at 07:01
  • @mlk I will look up those theorems. But can you tell me in the meantime why we can't use continuity of the norms to switch limits and integrals. I always though we could that, i.e. by the continuity of the norm we have $\lim_{n\to \infty}||f_n|| = ||\lim_{n\to \infty}f_n||$? This is a property of all continuous functions! – ManUtdBloke Sep 01 '17 at 07:03
  • @eurocoder If $f_n\to f$ in $L^p$-norm, then $|f_n|\to|f|$. But pointwise convergence does not imply convergence in $L^p$-norm (nor vice versa). – Angina Seng Sep 01 '17 at 20:26
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In general $ \lim_{n\to \infty} \int_\Omega |f_n(x)-f(x)|^p dx = \int_\Omega |\lim_{n\to \infty} f_n(x)-f(x)|^p dx$ is false !

Example: Let $p=1$ , $ \Omega =[0,1]$ and let $f_n$ be defined as follows $( n \ge 3)$:

$f_n(x)=n^2x$ , if $0 \le x \le 1/n$, $f_n(x)=-n^2x+2n$, if $1/n \le x \le 2/n$ and $f_(x)=0$, if $2/n \le x \le1$.

Fred
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  • But isn't the norm always a continuous function? And hence $\lim_{n\to \infty}||f_n|| = ||\lim_{n\to \infty}f_n||$? – ManUtdBloke Sep 01 '17 at 07:01