- #36

benorin

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Sorry it was late and I wanted to post it before bed... I'll show my work this time :)I think you made an integration error, but I'm not sure since you haven't shown what you've done. And what if ##q=4##?

**1.**Let ##1<p<4## and ##f\in L^p((1,\infty))## with the Lebesgue measure ##\lambda##. We define ##g\, : \,(1,\infty)\longrightarrow \mathbb{R}## by

$$

g(x)=\dfrac{1}{x}\int_x^{10x}\dfrac{f(t)}{t^{1/4}}\,d\lambda(t).

$$

Show that there exists a constant ##C=C(p)## which depends on ##p## but not on ##f## such that ##\|g\|_2 \leq C\cdot \|f\|_p## so ##g\in L^2((1,\infty)).## (FR)

**Work**: Let ##\tfrac{1}{p}+\tfrac{1}{q}=1## and note that ##1<p<4\Rightarrow \tfrac{4}{3}<q<\infty##.

$$\begin{gathered} \| g \| _2^2 =\int_1^\infty\left| \dfrac{1}{x}\int_x^{10x}\dfrac{f(t)}{t^{1/4}}\, d\lambda(t)\right| ^2\, d\lambda(x) \\ \leq \int_1^\infty \dfrac{1}{x^2} \left\{ \int_x^{10x} | f | ^p\, d\lambda(t)\right\} ^{\tfrac{2}{p}}\left\{\int_x^{10x} t^{-\tfrac{q}{4}}\, d\lambda(t) \right\} ^{\tfrac{2}{q}}\, d\lambda(x) \\ \end{gathered} $$

by Holder's inequality, also note that ##\forall x\in (1,\infty ),\quad \left\{ \int_x^{10x} | f | ^p\, d\lambda(t)\right\} ^{\tfrac{2}{p}}\leq \| f \| _p^2## hence it follows that

$$\begin{gathered} \| g \| _2^2 \leq \| f \| _p^2 \int_1^\infty \dfrac{1}{x^2} \left\{\int_x^{10x} t^{-\tfrac{q}{4}}\, d\lambda(t) \right\} ^{\tfrac{2}{q}}\, d\lambda(x) \\ = \| f \| _p^2 \int_1^\infty \dfrac{1}{x^2} \left\{\left[ \tfrac{t^{1-\tfrac{q}{4}}}{1-\tfrac{q}{4}}\right|_{t=x}^{10x} \right\} ^{\tfrac{2}{q}}\, d\lambda(x) \\ =\tfrac{4}{4-q} \| f \| _p^2 \int_1^\infty\left\{ \dfrac{1}{x^2} \left[ (10x)^{\tfrac{4-q}{4}}-x^{\tfrac{4-q}{4}}\right] ^{\tfrac{2}{q}} \right\} \, d\lambda(x) \\ =\tfrac{4}{4-q}\left( 10^{\tfrac{4-q}{4}}-1\right)^{\tfrac{2}{q}} \| f \| _p^2 \int_1^\infty x^{\tfrac{4-q}{4}\cdot \tfrac{2}{q}-2} \, d\lambda(x) \\ =\tfrac{4}{4-q}\cdot \tfrac{2q}{4-3q}\left( 10^{\tfrac{4-q}{4}}-1\right)^{\tfrac{2}{q}} \| f \| _p^2\underbrace{ \left[ x^{\tfrac{4-3q}{2q}}\right| _{x=1}^\infty}_{=-1} \\ =\tfrac{8q}{(4-q)(3q-4)}\underbrace{\left( 10^{\tfrac{4-q}{4}}-1\right)^{\tfrac{2}{q}}}_{\rightarrow 0 \text{ as }q\rightarrow 4} \| f \| _p^2 \\ \end{gathered} $$

**Edit**: I should've said something about this indeterminate form ##\tfrac{0}{0}## and l'Hospital's rule here, but @fresh_42 doesn't need the explanation of how to compute the limit,

*he knows.*

and the desired result comes upon taking square roots of both sides a remembering that ##q=q(p)## (q may be thought of as a function of ##p##).