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Let $(0,\infty)$ be equipped with the Lebesgue measure, and let $1 < p < \infty$. For each $ f \in L_{p}(\lambda)$ let $$T(f)(x) = x^{-1} \int f \chi_{(0,x)} \ d\lambda \quad \text{for} \ x >0$$

Then prove that $T$ is a bounded linear operator from $L_{p}(0,\infty)$ into itself.

4 Answers 4

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One (direct) way to see this is that $T$ is dominated by (a slightly modfied version of) the Hardy-Littlewood maximal operator, and it is a theorem that this is a bounded operator on $L^p$ for $p$ greater than 1. Or, directly, one can see that $T$ is bounded on $L^\infty$ and weak-bounded on $L^1$ (because if $f$ belongs to $ L^1$ then $Tf(x) <= ||f||_1/x$), so the Marcinkiewicz interpolation theorem applies. This is really much too indirect, though, but I'd like to point out that this is a special case of something more general. A direct approach is in page 242 of Inequalities (the page is available).

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It's a classic exercise, see for example Rudin's Real and Complex analysis (3rd ed.), exercise 14, chap.3, p. 72. There are detailed step-by-step hints, you should be fine.

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    Thanks Malik, i didnt know that!2010-08-18
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EDIT: This is just an approach, it doesn't actually work. I don't know if it's possible to make it correct.

Let $q = \frac{1}{1 - \frac{1}{p}}$ so $\frac{1}{p} + \frac{1}{q} = 1$.

By Hölder's inequality we have $$ \int_{0}^{x} |f(\lambda)| d\lambda \le (\int_{0}^{x} |f(\lambda)|^{p} d\lambda)^{\frac{1}{p}} (\int_{0}^{x} |1|^{q} d\lambda)^{\frac{1}{q}} \le ||f||_{p} x^{\frac{1}{q}} $$

Hence

$$ |\frac{1}{x} \int_{0}^{x} f(\lambda) d\lambda|^p \le |\frac{1}{x} \int_{0}^{x} |f(\lambda)| d\lambda|^p \le ||f||^{p}_{p} x^{\frac{1}{q}-1} $$

Now norm of $T(f)$ has the following form:

$$ ||T(f)||_{p} = (\int _{0}^{\infty} |T(f)(x)|^{p}dx)^{1/p} = (\int _{0}^{\infty}|\frac{1}{x} \int _{0}^{x} f(\lambda) d\lambda|^pdx)^{1/p} $$

Let $||f||^{p} = c$.

So

$$ ||T(f)||_{p} \le (\int _{0}^{\infty} c^{p} x^{\frac{1}{q}-1} dx)^{1/p} = ||f||_p (\int _{0}^{\infty} x^{-1/p})^{p} $$

Since $(\int _{0}^{\infty} x^{-1/p})^{p}$ is some constant we have $||T(f)|| _{p} \le ||f|| _{p} const$, hence this operator is bounded.

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    Wonderful! Solution.2010-08-17
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    It is not true that ( $\int_0^\infty x^{-1/p}$) is finite: in fact, it diverges for all $p$.2010-08-17
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    @Akhil Mathew: Yes, you are right. It seems that it's not that straightforward :)2010-08-17
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Following falagar's solution one can also get the value of $ \displaystyle ||T|| = \frac{p}{p-1}$