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Title says most of it. Could you help me find an example?

It is easy obviously to show a function that would not be in $L^p[0,1]$ for a specific $p$ (say $(1/x)^{1/p}$, but I can't see how it would be done for all $p$. The reason I'm asking is because we proved in class that $L^p[0,1]$ is nowhere dense as a subset of $L^1[0,1]$, so there must be some $L^1[0,1]$ like this..

Thanks :)

Added: thanks for all the comments. there was some missing parts about how to use convergence theorems that i couldn't complete my own so i'd love assistance :)

MathManiac
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  • $C[0,1]$ is dense in $L^1[0,1]$ and $C[0,1] \subset L^p[0,1]$. Thus $L^p$ is dense in $L^1$ –  Nov 27 '14 at 10:02
  • related MSE question http://math.stackexchange.com/questions/1040000/find-a-sequence-converging-to-zero-but-not-the-elemet-of-lp-space-for-every-1-p – Mirko Nov 27 '14 at 14:14

3 Answers3

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Take $$f(x) = a_n \quad\text{ if }\quad x\in\left(\frac{1}{2^n},\frac{1}{2^{n-1}}\right]$$ for $n = 1, 2, \dots$ and for some $a_n$.

You'll get $$\int_0^1 f(x)\, dx = \sum_{n=1}^{\infty}\frac{a_n}{2^n}$$ $$\int_0^1 f^p(x)\, dx = \sum_{n=1}^{\infty}\frac{a_n^p}{2^n}$$

Then choose the sequance $a_n$ so that the first sum is convergent, while the second one is divergent for any $p>1$. For example $$a_n = \frac{2^n}{n^2}.$$

$$\sum_{n=1}^{\infty}\frac{a_n}{2^n} = \sum_{n=1}^{\infty}\frac{1}{n^2}<\infty$$ $$\sum_{n=1}^{\infty}\frac{a_n^p}{2^n} = \sum_{n=1}^{\infty}\frac{2^{n(p-1)}}{n^{2p}} = \infty$$

Mher
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  • Thanks, that looks like a lovely solution. Just one question: how could I justify taking an infinite sum instead of finite? (the function can not be quite defined on the intervals if $n=\infty$. do I need to use some convergence theorem? – MathManiac Nov 27 '14 at 12:46
  • if I'd define $f_N(x)$ as you said, i'd get $\int_0^1 f_N(x)=\sum_{n=0}^N a_n/2^n$ which is good since $\int\lim f_N=\sum_{n=0}^\infty a_n/2^n$. but why $\lim f_N(x)\in L^1[0,1]$? why does it converge? – MathManiac Nov 27 '14 at 13:23
  • If the intervals do not overlap, then actually it doesn't matter how many the intervals are. But I guess you need the following: I used the countable additivity of Lebesgue integral, when I write $$\int_0^1 f(x),dx = \sum_{n=1}^{\infty}\int_{\left(\frac{1}{2^n},\frac{1}{2^{n-1}}\right]}f(x),dx = \sum_{n=1}^{\infty}\frac{a_n}{2^n}$$ – Mher Nov 27 '14 at 13:27
  • I actually did understand the part about the countable additivity. my problem is that you didn't quite defined f as you could tell the integral is an infinite sum like that with n->inf, since you only defined $f_N$ for finite $N$'s. I didn't understand why it doesn't matter how many intervals are – MathManiac Nov 27 '14 at 13:30
  • ok, let say as follows, define $f_n(x)=a_n$ when $x\in\left(\frac{1}{2^n},\frac{1}{2^{n-1}}\right]$ and $f_n(x)=0$ elsewhere. Then $f(x)=\sum_{n=1}^{\infty}f_n(x).$ Now, if you understand the part of countable additivity, then should get the point. – Mher Nov 27 '14 at 13:37
  • That's better, thanks! but one last thing - why $\sum f_n$ converges? (why f is well defined?) – MathManiac Nov 27 '14 at 13:44
  • ah, it that a problem at first !?..... that sum converges, because if you fix some $x\in(0,1]$ then there is only one $f_n$ for which $f_n(x)\ne 0$, for other $n$'s it is just zero. – Mher Nov 27 '14 at 13:46
  • Does it make sense for you ? – Mher Nov 27 '14 at 13:53
  • perfect, now everything is great. I understood it, and it's quite nice :)) thanks – MathManiac Nov 27 '14 at 15:08
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$$f = \sum_n \frac{1}{2^n \cdot \int_0^1 x^{-1+1/n} \, dx} \cdot x^{-1 + 1/n}$$

EDIT: The idea is that $x^{-1 + 1/n}$ is "barely integrable", but not integrable if taken to a power $p > \frac{-1}{-1 + 1/n} \to 1$. Hence, we sum all these functions in such a way that the result is still integrable (use monotone convergence).

But because of $f \geq C_n \cdot x^{-1 + 1/n}$ for all $n$ with $C_n > 0$, it is easy to see that $f^p$ is not integrable for any $p>1$.

PhoemueX
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How about this one? $$f(x)=\frac1{x(1-\ln(x))^2}$$ Then $$\int f(x)\,dx=\frac1{1-\ln x},$$ and it quickly follows that $f\in L^1$.

If $p>1$, pick $r$ with $p>r>1$ and write $$f(x)^p=\frac1{x^r}\cdot\frac1{x^{p-r}(1-\ln(x))^{2p}},$$ note that the first factor is non-integrable and the second factor is bounded below (it is continuous and positive on $(0,1]$ and unbounded near $x=0$), so $f(x)^p$ is non-integrable.

Harald Hanche-Olsen
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  • Thanks! could you explain more explicitly why $f(x)^p$ isn't integrable? I didn't quite understand the part aobut the second factor – MathManiac Nov 27 '14 at 13:20
  • The denominator of the second factor can be written as $(x^a(1-\ln x))^{2p}$ with $a=(p-r)/(2p)>0$. It's an easy calculus exercise to show that $x^a(1-\ln x)\to0$ when $x\to0$, so the second term (call it $s(x)$) goes to $\infty$ when $a\to0$. In particular, there is a $\delta>0$ so that $s(x)>1$ when $x<\delta$. On the compact interval $[\delta,1]$, $s$ is continuous, so it achieves a minumum there. It follows that $s(x)\ge b$ for some constant $b>0$, and all $x\in(0,1]$. And so $f(x)^p\ge b/x^r$, and we're done. – Harald Hanche-Olsen Nov 27 '14 at 17:38