Question about equivalence of norms of Banach space

Question

This is from "Excercise 24, "Stein-Shakarchi's Functional analysis"(Chapter 1)
we have $1\leq p<q\leq q <\infty$
Consider the Banach space $L^p\cap L^q$ with norm $\|f\|_{L^p\cap L^q}$=$\|f\|_{L^p}+\|f\|_{L^q}$ and let
$\Phi(t)=t^p,\text{ when }0\leq t \leq 1 \text{ and } = t^q \text{ When }1\leq t <\infty$
we have to show that $L^{\Phi}$ with its norm is equivalent to the space $L^p\cap L^q$
Let me define the Orlicz space $L^{\Phi}$.
Let $(X,\mu)$ be a measure space and $\Phi(t)$ is a continuous, convex and increasing function on $[0,\infty)$ with $\Phi(0)=0$ define $$L^{\Phi}=\{f:X\to \mathbb{R} \text{ such that } \int_{X}\Phi\left(\frac{|f(x)|}M\right)\,d\mu<\infty \text{ for some } M>0\}$$ and $$\|f\|_{\Phi}=\inf\{M>0\text{ such that }\int_{X}\Phi\left(\frac{|f(x)|}M\right)\,d\mu\leq 1\}.$$ I am not getting how to start proving it, please any hint or anything will help me, thanks.

Answer 1

To show that the norms are equivalent means you need to show that there exists constants $0 < A,B < \infty$ such that for any $f \in L^p \cap L^q$ it holds $$A||f||_{\Phi} \leq ||f||_{L^p \cap L^q} \leq B||f||_{\Phi}.$$

So fix an $f \in L^p \cap L^q$. Decompose $f$ into $f_0,f_1$ via \begin{align} f_0(x) &= f(x) \cdot \chi_{|f(x)| \leq \|f\|_{L^p\cap L^q}} \\ f_1(x) &= f(x) \cdot \chi_{|f(x)| > \|f\|_{L^p \cap L^q}} \end{align} so that $f = f_0 + f_1$. Note that $|f_0|, |f_1| \leq |f|$ so they both also belong to $L^p \cap L^q$ Using this decomposition we have by the sub-additivity of norms \begin{align} \|f\|_\Phi = \|f_0 + f_1\|_\Phi \leq \|f_0\|_\Phi + \|f_1\|_\Phi. \end{align} Now let's look at the two terms separately. For the first we have \begin{align} \int_X \Phi \left ( \frac{|f_0(x)|}{\|f\|_{L^p \cap L^q}}\right ) d\mu &= \int_X \Phi \left ( \frac{|f(x)\chi_{|f(x)| \leq \|f\|_{L^p \cap L^q}}|}{\|f\|_{L^p \cap L^q}}\right ) d\mu & (\text{Def of } f_0) \\ &= \int_X \left ( \frac{|f(x)\chi_{|f(x)| \leq \|f\|_{L^p \cap L^q}}|}{\|f\|_{L^p \cap L^q}} \right )^p d\mu & (*) \\ &= \int_X \left ( \frac{|f(x)|}{\|f\|_{L^p \cap L^q}} \right )^p\chi_{|f(x)| \leq \|f\|_{L^p \cap L^q}} d\mu \\ &\leq \int_X \left ( \frac{|f(x)|}{\|f\|_{L^p}} \right )^p d\mu & (\|f\|_{L^p} \leq \|f\|_{L^p \cap L^q})\\ &= 1 & (\text{Def of } \|f\|_{L^p}) \end{align} The line $(*)$ uses the fact that either the $|f(x)| \leq \|f\|_{L^p \cap L^q}$ and the whole argument is less than 1, or $|f(x)| > \|f\|_{L^p \cap L^q}$ and the whole argument is 0, which is still less than 1.

Note that we have just shown by definition that $\|f_0\|_{\Phi} \leq \|f\|_{L^p \cap L^q}$ since this is a valid choice of $M$ in the infimum defining $\|f_0\|_{\Phi}$.

In essentially the same manner we have \begin{align} \int_X \Phi \left ( \frac{|f_1(x)|}{\|f\|_{L^p \cap L^q}}\right ) d\mu &= \int_X \Phi \left ( \frac{|f(x)\chi_{|f(x)| > \|f\|_{L^p \cap L^q}}|}{\|f\|_{L^p \cap L^q}}\right ) d\mu & (\text{Def of } f_1) \\ &= \int_X \left ( \frac{|f(x)\chi_{|f(x)| > \|f\|_{L^p \cap L^q}}|}{\|f\|_{L^p \cap L^q}} \right )^q d\mu & (**) \\ &= \int_X \left ( \frac{|f(x)|}{\|f\|_{L^p \cap L^q}} \right )^q\chi_{|f(x)| > \|f\|_{L^p \cap L^q}} d\mu \\ &\leq \int_X \left ( \frac{|f(x)|}{\|f\|_{L^q}} \right )^q d\mu & (\|f\|_{L^q} \leq \|f\|_{L^p \cap L^q})\\ &= 1 & (\text{Def of } \|f\|_{L^q}) \end{align} Here the $(**)$ line uses that either the argument is 0, or it exceeds 1. This is basically the reversal of the $(*)$ line above. Ultimately this shows that $\|f_1\|_{\Phi} \leq \|f\|_{L^p \cap L^q}$ since this is a valid choice of $M$ in the infimum defining $\|f_1\|_{\Phi}$.

Combining these facts with the one above we have \begin{align} \|f\|_{\Phi} & \leq \|f_0\|_{\Phi} + \|f_1\|_{\Phi} \\ &\leq \|f\|_{L^p \cap L^q} + \|f\|_{L^p \cap L^q} \\ \implies \frac{1}{2} \|f\|_{\Phi} &\leq \|f\|_{L^p \cap L^q} \end{align} which shows the left inequality with $A = 1/2$.

For the other bound note that the function $\Phi$ satisfies for all $t \geq 0$ $$ \Phi(t) = \begin{cases} t^p & 0 \leq t \leq 1 \\ t^q & 1 < t \end{cases} \geq \begin{cases} t^q & 0 \leq t \leq 1 \\ t^q & 1 < t \end{cases} = t^q $$ since $1 \leq p < q$ implies $t^p \geq t^q$ for every $t \in [0,1]$

From this it follows \begin{align} \int_X \Phi \left ( \frac{|f(x)|}{\|f\|_{L^q}} \right ) d\mu &\geq \int_X \left ( \frac{|f(x)|}{\|f\|_{L^q}} \right )^q d\mu \\ &= 1 \end{align} and the equality holds only if $|f(x)| \geq \|f\|_{L^q}$ for $\mu$ a.e. $x$. This is enough to show that $\|f\|_\Phi \geq \|f\|_{L^q}$ (a fact which you should spend a moment convincing yourself of to reinforce your understanding).

It also always holds that for $1 \leq p \leq q$ that $\|f\|_{L^p} \leq \|f\|_{L^q}$. Using this and the fact above we have \begin{align} 2\|f\|_\Phi &= \|f\|_\Phi + \|f\|_{\Phi} \\ &\geq \|f\|_{L^q} + \|f\|_{L^q} \\ &\geq \|f\|_{L^p} + \|f\|_{L^q} \\ &= \|f\|_{L^p \cap L^q} \end{align} which shows the right inequality with $B = 2$.