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Is the sequence $(x_n)$ convergent in the space $L^1(0,1)$ ?

$x_n(t)= n^2 t^n (1-t^2)$ for $n\in\mathbb{N}$.

norm: $\|x\|=\int_{(0,1)} \left|x(t)\right| \; dt$

I think it should converge to zero, by checking the graph. but I cannot show it explicitly.

Could you please help?

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Have you tried using dominated convergence? As far as i can see, this will give you the desired convergence directly. –  Ukhrir Feb 27 at 12:09
    
Note $x_n\ge0$ and $(x_n)$ converges pointwise to the zero function. You can compute the required integrals explicitly, and easily. –  David Mitra Feb 27 at 12:20
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2 Answers 2

up vote 2 down vote accepted
  • $\left| (x_n)(t) \right|$ converges pointwise to the $0$ function on $[0,1]$.

  • We have \begin{align*} \left\|{x_n}\right\| &= \int_0^1 \left(n^2 t^n - n^2 t^{n+2}\right) \; dt \\ &= \frac{n^2}{n+1} - \frac{n^2}{n + 3} \\ &= \frac{2n^2}{(n+1)(n+3)} \to 2 \\ \end{align*}

Since $L^1$ convergence implies pointwise convergence almost everywhere for some subsequence, the only candidate for $x_n$ to converge to in the $L^1$ norm is the zero function. But $\left\|{x_n - 0}\right\| \to 2 \ne 0$, so $x_n$ does not converge to any function in the $L^1$ norm.


Direct proof that $L^1$ convergence implies pointwise convergence almost everywhere for some subsequence:

Note: This proof had a subtle error before, which has now been fixed.

Let $f_n \to f$ in $L^1$. Then for any $k$, let $E_{n,k}$ be the set of points where $\left|f_n - f\right| > \frac{1}{k}$. Note that $\left\|{f_n - f}\right\|_{1} \ge \frac{1}{k} \mu(E_{n,k})$, implying that for a fixed $k$, $\mu(E_{n,k}) \to 0$.

For each $i$, choose an integer $n_i$ so that $n_i$ is an increasing sequence of integers, and $\mu(E_{n_i, i}) < \frac{1}{2^i}$. Note that this means $\mu(E_{n_i, k}) < \frac{1}{2^i}$ whenver $k \le i$.

Now consider the set $E$ of points $x$ where $f_{n_i}(x) \not \to f(x)$. We have $f_{n_i}(x) \not \to f(x)$ if and only if there is some $k$ such that $\left|f_{n_i}(x) - f(x) \right| > \frac1k$ infinitely often. Therefore, $$ E = \bigcup_{k=1}^\infty \bigcap_{N=1}^\infty \bigcup_{i=N}^\infty E_{n_i,k} $$ So \begin{align*} \mu(E) &\le \sum_{k=1}^\infty \lim_{N \to \infty} \sum_{i=N}^\infty \mu(E_{n_i,k}) \\ &= \sum_{k=1}^\infty \lim_{N \to \infty}_{N \ge k} \sum_{i=N}^\infty \mu(E_{n_i,k}) \\ &\le \sum_{k=1}^\infty \lim_{N \to \infty}_{N \ge k} \sum_{i=N}^\infty \mu(E_{n_i,i}) \quad \quad (\text{since } k \le i) \\ &\le \sum_{k=1}^\infty \lim_{N \to \infty}_{N \ge k} \sum_{i=N}^\infty \frac{1}{2^{i}} \\ &= \sum_{k=1}^\infty (0) = 0\\ \end{align*}

Therefore, $f_{n_i} \to f$ pointwise outside the set $E$ of measure $0$.


In fact, the above proof used the fact that $f_n \to f$ in measure, which is a weaker (more general) condition than convergence in $L^1$ norm. For an extended discussion of this fact (Corollary 3, Exercise 6) and others, see Terry Tao's blog post. See also this more recent mathSE question, in which I referred to this answer and spotted the error in the proof.

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thanks. could you please explain the part: "$L^1$ convergence implies pointwise convergence almost everywhere" –  104078 Feb 27 at 16:28
    
@user104078 I didn't type that quite right; I meant $L^1$ convergence implies pointwise convergence almost everywhere for some subsequence. –  Goos Feb 27 at 17:00
    
@user104078 I've added a proof of that fact (which was stated incorrectly before). Let me know if that helps. –  Goos Feb 27 at 17:16
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$\{f_n\}$ does not converge in the $L^1$ sense.

Observation I. $\lim_{n\to\infty}\int_0^1 f_n(x)\,dx=2.$

Observation II. For every $0<a<1$, the sequence $f_n$ converges uniformly to $0$ in $[0,a]$.

So, due to II, for every $a\in(0,1)$, there exists an $n_0\in\mathbb N$, such that for $n\ge n_0$: $$ \int_a^{1}f_n(x)\,dx>3/2 \quad\text{and}\quad \int_0^a f_n(x)\,dx<1/2. $$ Then, due to I, we can find an $m>n$, such that $f_m(x)<1/2$, for $x\in[0,a_n]$, and thus $\int_{a_n}^1 f_m(x)\,dx>3/2$. Thus $$ \int_0^1 \lvert f_m-f_n\rvert\,dx\ge\int_{a_n}^1 \lvert f_m-f_n\rvert\,dx\ge \int_{a_n}^1 f_m-\int_{a_n}^1 f_n\ge \frac{3}{2}-\frac{1}{2}=1. $$

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"For every $n$, there exists an $a_n < 1$, such that $\int_0^{a_n} f_n(x) \; dx > 3/2$" is false. For $n = 1$, we have $\int_0^1 f_1 = \int_0^1 t(1 - t^2) \; dt = \frac14$, so no subinterval has an integral bigger than $1/4$. –  Goos Feb 27 at 12:40
    
@Goos: Correct. I have made the appropriate changes. –  Yiorgos S. Smyrlis Feb 27 at 13:18
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