Find limit of sequence of functions

**wglmb** · November 4th 2009, 09:50 AM

Let $f_n\in C[0,1]$ be defined by
$f_n =\begin{cases}n^{2/3} & \mbox{if }0\le x\le \frac{1}{n}\\2n^{2/3}-n^{5/3}x & \mbox{if }\frac{1}{n} < x< \frac{2}{n}\\0 & \mbox{if }\frac{2}{n} \le x\le 1\\\end{cases}$

Let $\parallel f \parallel _1 = \int_{0}^{1}\mid f(x)\mid\, dx$ and $\parallel f \parallel _2 = \left (\int_{0}^{1}\mid f(x)\mid ^2\, dx \right )^\frac{1}{2}$

Show that $f_n \rightarrow 0$ with respect to $\parallel \cdot \parallel _1$ , but $f_n$ is not convergent with respect to $\parallel \cdot \parallel _2$

---------------------

Now, I did the following, but it must be wrong since it doesn't depend on the choice of norm, so I get that $f_n \rightarrow 0$ in both cases.

Choose $x \in [0,\tfrac{1}{n}]$ for some small n.
$\exists \, n_0 > n$ st $x \notin [0,\tfrac{1}{n} ]$ but rather $x \in (\tfrac{1}{n} ,\tfrac{2}{n} )$
$\exists \, \tilde{n}_0 > n_0$ st $x \notin (\tfrac{1}{n},\tfrac{2}{n} )$ but rather $x \in [\tfrac{2}{n} ,1 )$
$\Rightarrow \, \exists \, \hat{n}_0$ st $f_n = 0 \, \forall \, n \ge \hat{n}_0$

So, clearly, we have that $\forall \, \epsilon >0 \, \exists \, \hat{n}_0 \in \mathbb{N}$ st $\parallel 0 - f_n \parallel _1 < \epsilon \, \forall \, n \ge \hat{n}_0$

So what have I missed that stops the above from working, and what is a better way of doing it?

**putnam120** · November 4th 2009, 07:49 PM

[mistaken]

**Laurent** · November 5th 2009, 12:36 AM

Originally Posted by wglmb

Let $f_n\in C[0,1]$ be defined by
$f_n =\begin{cases}n^{2/3} & \mbox{if }0\le x\le \frac{1}{n}\\2n^{2/3}-n^{5/3}x & \mbox{if }\frac{1}{n} < x< \frac{2}{n}\\0 & \mbox{if }\frac{2}{n} \le x\le 1\\\end{cases}$

Let $\parallel f \parallel _1 = \int_{0}^{1}\mid f(x)\mid\, dx$ and $\parallel f \parallel _2 = \left (\int_{0}^{1}\mid f(x)\mid ^2\, dx \right )^\frac{1}{2}$

Show that $f_n \rightarrow 0$ with respect to $\parallel \cdot \parallel _1$ , but $f_n$ is not convergent with respect to $\parallel \cdot \parallel _2$

Why not just compute $\|f_n\|_1$ and $\|f_n\|_2$ ? You can notice that $f_n(x)\geq 0$ hence

$\int |f_n(x)|^p dx =( n^{2/3})^p \frac{1}{n}+\int_{1/n}^{2/n} (n^{2/3}-2n^{5/3}x)^p dx$

for $p=1,2$ . For $p=1$ , if you get 0, you're done. For $p=2$ , you should not get 0; if you get an infinite limit, there is no limit in $L^2$ . If there is a finite limit... it may need some more work to find it or prove there is none. (I think here you get an infinite limit, just because of the first term $(n^{2/3})^p \frac{1}{n}$ , so you're done)

In fact, since the space is $[0,1]$ , if $f_n\to f$ in $L^2$ , then $f_n\to f$ in $L^1$ hence the limit should be 0 and it suffices to check if $\lim_n \|f_n\|_2=0$ to know if there is convergence in $L^2$ .

Now, I did the following, but it must be wrong since it doesn't depend on the choice of norm, so I get that $f_n \rightarrow 0$ in both cases.

Choose $x \in [0,\tfrac{1}{n}]$ for some small n.
$\exists \, n_0 > n$ st $x \notin [0,\tfrac{1}{n} ]$ but rather $x \in (\tfrac{1}{n} ,\tfrac{2}{n} )$
$\exists \, \tilde{n}_0 > n_0$ st $x \notin (\tfrac{1}{n},\tfrac{2}{n} )$ but rather $x \in [\tfrac{2}{n} ,1 )$
$\Rightarrow \, \exists \, \hat{n}_0$ st $f_n = 0 \, \forall \, n \ge \hat{n}_0$

So, clearly, we have that $\forall \, \epsilon >0 \, \exists \, \hat{n}_0 \in \mathbb{N}$ st $\parallel 0 - f_n \parallel _1 < \epsilon \, \forall \, n \ge \hat{n}_0$

What you proved here is that $f_n$ converges to $0$ almost-everywhere (in fact, everywhere). As you probably know, you may have $f_n(x)\to f(x)$ for almost all (or all) $x$ , and $\int f_n \not\rightarrow \int f$ . This is however true under a few conditions: cf. the bounded convergence theorem. In this case it is not straightforward to apply it to prove convergence in $L^1$ , and anyway it wouldn't suffice for the $L^2$ case.

**wglmb** · November 5th 2009, 03:33 AM

Thanks very much, that was great help!

wglmb

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