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I have to show that $f(x) \neq \lim_{n\to\infty} f_n(x_n) $, where $x_n$ is a converging sequence in $[0,1]$, and $f_n \rightarrow f$ in the $L^1$ norm, such that $ \int_0^1 | f_n (x) - f(x) | dt < \epsilon $.

So, as far as I got, I need to find a function which does not converge pointwise, while it converges in the norm. Getting examples from uniform vs pointwise convergence is quite easy, but I cannot work this out: it's like if I have to come out with a function whose integral e.g. goes to zero but the function moves around in the interval.

I'm looking for a countrexample to something like stated here.

Mino
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2 Answers2

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Take for example $f_n(x)=x^n$. Then $f_n\to f=0$ in $L^1([0,1])$: $$\int_0^1|f_n(x)-f(x)|dx=\int_0^1 x^n dx=\frac{1}{n+1}\to 0.$$ On the other hand, $x_n=1-\frac{1}{n^2}\to 1$ and $$\lim_{n\to\infty}f_n(x_n)=\lim_{n\to\infty}\left(1-\frac{1}{n^2}\right)^n=1\not=f(1)=0.$$

Robert Z
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  • so it is sufficient to get a discoutinuous limit and let the $x_n$ sequence go to the "other" limit? My doubt is on the L^1 norm: could you please show me why for that solution the L^1 inequality holds? thank you so much – Mino Sep 11 '16 at 21:11
  • Sure. I edited my answer. – Robert Z Sep 11 '16 at 21:13
  • great: so we you calulate the integral how do you select $f(x)$ in general? I know this is a very basic quastion but I want to be sure I understand – Mino Sep 11 '16 at 21:17
  • Sorry no my bad, I mean where you plug in the final 1 – Mino Sep 11 '16 at 21:24
  • no, why $f(1) = 0$, which is the function you're considenring. i'm overwhelmed I'm a bit confused. the rest is shining clear – Mino Sep 11 '16 at 21:27
  • oh maybe I see. you have to plug in in the $lim$ of $f_n(x)$. but since it's a constant (equals zero), no matter which x you put, it's always zero. is it right? in other cases it may change – Mino Sep 11 '16 at 21:30
  • @jack I defined $f$ to be the zero function in $[0,1]$. However a function in $L^1$ is determined up to a set of measure zero. So also the function which is zero everywhere and $1$ at $1$ is a good limit. In fact your question is a bit strange from this point of view... – Robert Z Sep 11 '16 at 21:31
  • can you explain me in more detail (or give me a reference, I don't want to take away your time) what does "a function in L1 is determined up to a set of measure zero" mean? – Mino Sep 11 '16 at 21:41
  • @Jack Take a look here: https://proofwiki.org/wiki/A.E._Equal_Positive_Measurable_Functions_have_Equal_Integrals – Robert Z Sep 12 '16 at 07:24
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Let $f_n = \Bbb e ^{-nx} : [0,1] \to \Bbb R$ and $x_n = \frac 1 n \to 0$. Let $f = \lim f_n$ with the limit understood in the $L^1$ norm. I claim that $f = 0$ because

$$\int \limits _0 ^1 |f_n (x) - 0| \ \Bbb d x = \int \limits _0 ^1 \Bbb e ^{-nx} \ \Bbb d x = \frac {\Bbb e ^{-nx}} {-n} \Bigg| _0 ^1 = \frac {1 - \Bbb e ^{-n}} {n} \to 0 .$$

Notice now that $f_n (x_n) = \Bbb e^{-1} \not\to 0 = f(0)$.

To make things even funnier, notice that the limit of the sequence $(f_n) _{n \ge 1}$ in the usual topology (of pointwise convergence) is the function $g(x) = \begin{cases} 1, & x = 0 \\ 0, & x \in (0, 1] \end{cases} .$

Alex M.
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