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 December 2nd, 2012, 04:12 PM #1 Newbie   Joined: Dec 2012 Posts: 9 Thanks: 0 LEBESGUE INTEGRATION Does anyone know a function belonging to L^p, but not to L^q, for all qp?. "p" is a real number which is arbitrary but fixed. I think these functions exist, but I am not able to find one of them. Thanks.
 December 3rd, 2012, 01:27 PM #2 Global Moderator   Joined: May 2007 Posts: 6,628 Thanks: 622 Re: LEBESGUE INTEGRATION You need to define the domain of integration.
 December 3rd, 2012, 03:04 PM #3 Newbie   Joined: Dec 2012 Posts: 9 Thanks: 0 Re: LEBESGUE INTEGRATION Well, you are right, but I did not define it because you can choose the domain of integration, as long as it works.
 December 4th, 2012, 01:13 PM #4 Global Moderator   Joined: May 2007 Posts: 6,628 Thanks: 622 Re: LEBESGUE INTEGRATION Example 1/?x (interval [0,1]). Belongs to Lp for p < 2, not Lq for q ? 2. (interval [1,?). Belongs to Lp for p >2, not Lq for q ? 2.
 December 5th, 2012, 06:06 AM #5 Newbie   Joined: Dec 2012 Posts: 9 Thanks: 0 Re: LEBESGUE INTEGRATION That' s not exactly what I asked. I could explain it better, but I am not sure how to use mathematical symbols here. I know how to use LaTeX, however, I don't know how to use it here.
 December 5th, 2012, 10:49 AM #6 Math Team   Joined: Sep 2007 Posts: 2,409 Thanks: 6 Re: LEBESGUE INTEGRATION Begin with the HTML tag [ t e x ] and end with [ / t e x ]- without the spaces, of course.
 December 6th, 2012, 05:22 AM #7 Newbie   Joined: Dec 2012 Posts: 9 Thanks: 0 Re: LEBESGUE INTEGRATION Thanks for your help. I explain it better, Let $\lambda$ be Lebesgue measure on $\mathbb{R}$. Let $p \in (1,\infty)$. Find a function in $L_p(\lambda)$but that fails to be in $L_q(\lambda)$ for $q \in [1,p)$. Find another function in $L_p(\lambda)$ that fails to be in $L_q(\lambda)$ for $q \in (p,\infty).$
 December 6th, 2012, 02:05 PM #8 Global Moderator   Joined: May 2007 Posts: 6,628 Thanks: 622 Re: LEBESGUE INTEGRATION Just to be clear, the domain of integration is the whole real line? If so, just use the examples I gave where the functions = 0 outside the original intervals. The only difficulty is that Lp would be on the divergent side of the interval, i.e. divergent for q ? p or p ? q for these examples.
 December 7th, 2012, 07:13 AM #9 Newbie   Joined: Dec 2012 Posts: 9 Thanks: 0 Re: LEBESGUE INTEGRATION That's precisely the difficulty. For instance, if I choose the function $\frac{1}{x}$ defined on (0,1), that function does not belong to $L^1((0,1))$ but it belongs to $L^q((0,1))$, for all q>p. I was looking for the opposite property. I have finally found an example. Let $\{p_n +\}_{n=1}^{\infty}\subset (1,p)$ be a sequence converging to p. For every $n\in \mathbb{N}$ we can find a function $f_n$ such that, (a)$f_n \in L^p((1,\infty))$ and $f_n \notin L^q((1,\infty))$, for any q $\in (0,p_n +].$ (For example $\frac{1}{x^{\frac{1}{p_n +}}}$) (b)$||f_n||_p<\frac{1}{2^n}.$ (just multiply by a constant). We define, $f=\sum_{n=1}^{\infty} f_n$
 December 7th, 2012, 07:34 AM #10 Newbie   Joined: Dec 2012 Posts: 9 Thanks: 0 Re: LEBESGUE INTEGRATION That function is an example of a function belonging to $L^p((1,\infty))$ but not to $L^q((1,\infty))$, for any qp. Extending these 2 function to $(0,\infty)$ and considering their sum, we can even find a function belonging to $L^p((0,\infty))$ but not to $L^q((0,\infty))$, for any q different to p. PS: I wanted to edit my last post because I did not press submit intentionally but I don't know how.

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