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Problem of the Week #258 - Oct 17, 2017

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Euge

MHB Global Moderator
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Jun 20, 2014
1,892
Here is this week's POTW:

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Find the norm of the linear operator $T: \mathscr{L}^p(0, \infty) \to \mathscr{L}^p(0, \infty)$ defined by the equation

$$(Tf)(x) = \frac{1}{x}\int_0^x f(t)\, dt$$

Here it is assumed that $1 < p < \infty$.

Note: The space $\mathscr{L}^p(0,\infty)$ consists of all Lebesgue integrable functions $f : (0,\infty) \to \Bbb R$ such that $\|f\|_p < \infty$. For $p < \infty$, $\|f\|_p := \left(\int_0^\infty \lvert f(x)\rvert^p\, dx\right)^{1/p}$.
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Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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Euge

MHB Global Moderator
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Jun 20, 2014
1,892
No one answered this week's problem. You can read my solution below.


The norm is $q := p/(p-1)$, the exponent conjugate to $p$. Let $f\in L^p(0,\infty)$, and let $r\in (0, 1/q)$ where $r$ is a constant to be determined. Writing $f(t) =t^{-r}[f(t)t^{r }]$ and applying Hölder's inequality, we obtain

$$\lvert Tf(x)\rvert^p \le (1 - r q)^{-p/q} x^{-1-rp} \int_0^x \lvert f(t)\rvert^{p} t^{rp}\, dt$$

Integrating over $(0, \infty)$ and changing the order of integration, we find

$$\int_0^\infty \lvert Tf(x)\rvert^p \le (1 - rq)^{-p/q}(rp)^{-1} \int_0^\infty \lvert f(t)\rvert^p\, dt$$

Hence

$$\|Tf\|_p \le (1 - rq)^{-1/q}(rp)^{-1/p}\|f\|_p$$

Choosing $r$ such that $1 - rq = rp$, i.e., $r = \frac1{pq}$ (since $\frac{1}{p} + \frac{1}{q} = 1$),

$$\|Tf\|_p \le (rp)^{-1/q - 1/p} \|f\|_p = q\|f\|_p$$

It follows that $\|T\|\le q$. To obtain $\|T\| = q$, we choose the function $f : (0, \infty) \to (0, \infty)$, $f(x) = x^{-1/p}1_{[1,R]}(x)$, where $R$ is scaled sufficiently large.
 
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