# L^p spaces

• Mar 7th 2009, 02:54 AM
HenryB
L^p spaces
Let $\displaystyle p>1$. Give examples of sequences $\displaystyle (f_n)$ and $\displaystyle (g_n)$ in $\displaystyle L^{p} (0,1)$ such that

i) $\displaystyle lim_{n\rightarrow \infty} f_n(x) = 0$ almost everywhere but $\displaystyle lim_{n\rightarrow \infty} ||f_n||_{p} \neq 0$.

ii) $\displaystyle lim_{n\rightarrow \infty} ||g_n||_{p} = 0$ but $\displaystyle lim_{n\rightarrow \infty} g_n(x)$ does not exist for any $\displaystyle x \in (0,1)$.
• Mar 7th 2009, 03:45 AM
Opalg
Quote:

Originally Posted by HenryB
Let $\displaystyle p>1$. Give examples of sequences $\displaystyle (f_n)$ and $\displaystyle (g_n)$ in $\displaystyle L^{p} (0,1)$ such that

i) $\displaystyle lim_{n\rightarrow \infty} f_n(x) = 0$ almost everywhere but $\displaystyle lim_{n\rightarrow \infty} ||f_n||_{p} \neq 0$.

ii) $\displaystyle lim_{n\rightarrow \infty} ||g_n||_{p} = 0$ but $\displaystyle lim_{n\rightarrow \infty} g_n(x)$ does not exist for any $\displaystyle x \in (0,1)$.

For i), try $\displaystyle f_n(x) = n^{1/p}x^n$.

ii) is trickier. You need to use something like the Haar functions.
• Mar 7th 2009, 02:02 PM
HenryB
Thanks a lot. (i) is a good answer but is there really nothing simpler for (ii)? It doesn't seem like the sort of solution that you can really spot yourself.

The question goes on to say:

"For each $\displaystyle \epsilon > 0$ fnd a subset $\displaystyle E_\epsilon$ of [0,1] such that $\displaystyle f_n(x) \rightarrow 0$ uniformly on $\displaystyle [0,1]$ \ $\displaystyle E_\epsilon$.

Now it's clearly not what the question is after (I don't actually know what the question is trying to say) but can't we just take $\displaystyle E_\epsilon = (0,1]$ for all $\displaystyle \epsilon > 0$?

Secondly, can we find a subsequence $\displaystyle ({g_n}_r)$ such that $\displaystyle lim_{r \rightarrow \infty} {g_n}_r(x) = 0$ almost everywhere?
• Mar 8th 2009, 09:53 AM
Opalg
Quote:

Originally Posted by HenryB
is there really nothing simpler for (ii)? It doesn't seem like the sort of solution that you can really spot yourself.

The difficulty with (ii) is that to ensure $\displaystyle \|g_n\|_p\to0$ you want g_n to be small over most of the interval; but to prevent $\displaystyle \lim_{n\to\infty}g_n(x)$ existing at every point of the interval you need to arrange that, for every x, g_n(x) stays well away from 0 for infinitely many values of n. The only convenient way I know to achieve that is to define g_n to be 0 on most of the interval, but to bob up to 1 on a small subinterval that gradually works its way along the whole interval.

The easiest way to do this is to construct g_n for n in batches of length $\displaystyle 2^k$. For $\displaystyle n = 2^k+j$, where $\displaystyle 0\leqslant j<2^k$, define $\displaystyle g_n(x)$ to be 1 on the interval $\displaystyle \left[\frac j{2^k},\frac{j+1}{2^k}\right]$, and 0 on the remainder of the interval [0,1].

Quote:

Originally Posted by HenryB
The question goes on to say:

"For each $\displaystyle \epsilon > 0$ find a subset $\displaystyle E_\epsilon$ of [0,1] such that $\displaystyle f_n(x) \rightarrow 0$ uniformly on $\displaystyle [0,1]$ \ $\displaystyle E_\epsilon$.

Now it's clearly not what the question is after (I don't actually know what the question is trying to say) but can't we just take $\displaystyle E_\epsilon = (0,1]$ for all $\displaystyle \epsilon > 0$?

Secondly, can we find a subsequence $\displaystyle ({g_n}_r)$ such that $\displaystyle lim_{r \rightarrow \infty} {g_n}_r(x) = 0$ almost everywhere?

For the example $\displaystyle f_n(x) = n^{1/p}x^n$ that I gave, the only point at which $\displaystyle f_n(x)\not\to0$ is x=1. The sequence does not converge uniformly on [0,1], or even on [0,1), but it does converge to 0 uniformly on $\displaystyle [0,1-\varepsilon]$, for all $\displaystyle \varepsilon > 0$.

For the above sequence (g_n), the subsequence $\displaystyle (g_{2^k})$ converges to 0 almost everywhere (in fact, everywhere except at x=0).
• Mar 9th 2009, 04:49 AM
HenryB
Great help - thanks a lot indeed.