Question for the proof of $L^p$ space

Question

There is a theorem stated in my textbook as following and my question is below the proof :

Let $0<p<q<\infty$ and let $f$ in $L^{p,\infty}(X,\mu) \cap L^{q,\infty}(X,\mu)$, where $X$ is a $\sigma$-finite measure space.

Then $f$ is in $L^{r}(X,\mu)$ for all $r\in(p,q)$, where $L^{p,\infty}(X,\mu)$ is the well-known space of weak-$L^p(X,\mu)$.

Proof: Recall that $\omega(\alpha)=\mu\bigg(\{x\in X : |f(x)|>\alpha\}\bigg)$. Since $f$ is in $L^{p,\infty}(X,\mu) \cap L^{q,\infty}(X,\mu)$,we take $$\omega(\alpha)\leq \min\bigg\{\dfrac{\lVert f\rVert^p_{L^{p,\infty}}}{\alpha^p},\dfrac{\lVert f\rVert^q_{L^{q,\infty}}}{\alpha^q}\bigg\}.$$

Set

$$B=\bigg(\dfrac{\lVert f\rVert^q_{L^{q,\infty}}}{\lVert f\rVert^p_{L^{p,\infty}}}\bigg)^{\dfrac{1}{q-p}}.$$

We now estimate the $L^r$ norm of $f$ as follows:

\begin{align} \lVert f\rVert^r_{L^r} &=r\int_0^\infty \alpha^{r-1}\omega(\alpha)\,d\alpha\\ &\leq r\int_0^\infty \alpha^{r-1} \min\bigg\{\dfrac{\lVert f\rVert^p_{L^{p,\infty}}}{\alpha^p},\dfrac{\lVert f\rVert^q_{L^{q,\infty}}}{\alpha^q}\bigg\}\,d\alpha\\ &\large\color{red}=r\int_0^B \alpha^{r-1-p}\lVert f\rVert^p_{L^{p,\infty}} \, d\alpha + r\int_B^\infty \alpha^{r-1-q}\lVert f\rVert^q_{L^{q,\infty}} \, d\alpha\\ &=\dfrac{r}{r-p}\lVert f\rVert^p_{L^{p,\infty}}B^{r-p}+\dfrac{r}{q-r}\lVert f\rVert^q_{L^{q,\infty}}B^{r-q}\\ &=\bigg(\dfrac{r}{r-p} +\dfrac{r}{q-r}\bigg)\bigg(\lVert f\rVert^p_{L^{p,\infty}}\bigg)^{\dfrac{q-r}{q-p}}\bigg(\lVert f\rVert^q_{L^{q,\infty}}\bigg)^{\dfrac{r-p}{q-p}}. \end{align} Observe that the integrals converges since $r-p>0$ and $r-q<0$

I have no ideal for the red equality holding,so my question is:

Why in the interval $[0,B]$ the minimum is $\dfrac{\lVert f\rVert^p_{L^{p,\infty}}}{\alpha^p}$ and in the interval $[B,\infty)$ the minimum is $\dfrac{\lVert f\rVert^q_{L^{q,\infty}}}{\alpha^q} $?

Is there anyone can give me some hint for that I stuck with? Thanks for patiently reading and considering my request.

Answer 1

This is a classical technique in harmonic analysis with the title: let some constant to be determined later. You may try to put $B=1$ to see what happen, of course, you cannot conclude anything actually. So the strategy is to keep $B$ fixed and we will config the value of $B$ while needed.

If we want $\min\{\alpha^{-p}\|f\|_{p,\infty}^{p},\alpha^{-q}\|f\|_{q,\infty}^{q}\}=\alpha^{-p}\|f\|_{p,\infty}^{p}$, this means $\alpha^{-p}\|f\|_{p,\infty}^{p}\leq\alpha^{-q}\|f\|_{q,\infty}^{q}$, so $\dfrac{\|f\|_{q,\infty}^{q}}{\|f\|_{p,\infty}^{p}}\geq\dfrac{\alpha^{q}}{\alpha^{p}}=\alpha^{q-p}$, so $\alpha\leq B$, where we need the role of $B$ to be satisfying in this fashion.

Further note here: in elementary mathematical analysis, when we deal with integral like $\displaystyle\int_{0}^{\infty}\dfrac{e^{-x}}{x^{1/2}}dx$, one deal with random partition of the domain of the integral to be $[0,c]$ and $[c,\infty)$ and it still goes through: $\displaystyle\int_{0}^{c}\dfrac{e^{-x}}{x^{1/2}}dx+\int_{c}^{\infty}\dfrac{e^{-x}}{x^{1/2}}dx<\infty$. Unfortunately, when we deal with more delicate integral like your question, usually one cannot set $B$ to be arbitrary number, it depends heavily on the context.

Answer 2

This is simply $$ \frac{\|f\|_q^q}{\alpha^q}\geq\frac{\|f\|_p^p}{\alpha^p} \iff \alpha^{q-p}\leq \frac{\|f\|_q^q}{\|f\|_p^p} \iff \alpha\leq \left(\frac{\|f\|_q^q}{\|f\|_p^p}\right)^{\frac1{q-p}}=B. $$ Thus $$ \min\left\{\frac{\|f\|_q^q}{\alpha^q},\frac{\|f\|_p^p}{\alpha^p}\right\} =\begin{cases} \frac{\|f\|_p^p}{\alpha^p},&\ \alpha\leq B \\ \ \\ \frac{\|f\|_q^q}{\alpha^q},&\ \alpha>B \end{cases} $$