Equivalence between $L^{p_0}+ L^{p_1}$ and Orlicz space $L^\Psi$ with suitable $\Psi$

Question

This is from Exercise 24, Chapter 1, in Stein and Shakarchi's Functional Analysis. First a few definitions.

$L^{p_0}+L^{p_1}$ is defined as the vector space of measurable functions $f$ on a measure space $X$, that can be written as a sum $f=f_0+f_1$ with $f_0\in L^{p_0}$ and $f_1\in L^{p_1}$. Define $$\|f\|_{L^{p_0}+L^{p_1}}=\inf\{\|f_0\|_{L^{p_0}}+\|f_1\|_{L^{p_1}}\},$$ where the infimum is taken over all decomposition $f=f_0+f_1$ with $f_0\in L^{p_0}$ and $f_1\in L^{p_1}$.

Suppose $\phi(t)$ is continuous, convex, increasing function on$[0,\infty)$, with $\phi(0)=0$ and $\phi(t)$ is not the constant function 0. Define $$L^{\phi}=\{f\hspace{0.2cm} \text{measurable:}\int_{X}\phi(|f(x)|/M)d\mu<\infty, \text{for some}\hspace{0.2cm} M>0\}$$ and $||f||_{\phi}=\inf_{M>0}\int_{X}\phi(|f(x)|/M)d\mu\le1$.

Suppose $1\leq p_0<p_1<\infty$ and define $$\Psi(t) = \int_0^t \psi(u) du \quad \text{where} \; \psi(u) = \begin{cases} u^{p_1-1} & \text{if $u<1$} \\ u^{p_0-1} & \text{if $1\leq u<\infty$} \end{cases}$$

Show that $L^\Psi$ with its norm is equivalent to the space $L^{p_0} + L^{p_1}$. In other words, there exist $A, B > 0$, so that $$A \| f \|_{L^{p_0} + L^{p_1}} \leq \| f \|_{L^\Psi} \leq B \| f \|_{L^{p_0} + L^{p_1}} .$$

The left half of the inequality is easy to prove, as $\Psi(|f|/M)$ effectively defines a decomposition of $f$. I cannot figure out how to prove the right half of the inequality.

Answer 1

There is a tacit assumption that $1 \leq p_0 \leq p_1$. I'll include the reason why this is at the end.

Now let $f \in L^{p_0} + L^{p_1}$. Since we do not know if the infimum is attained we are going to need a bit of slack. So let $f_0 \in L^{p_0}, f_1 \in L^{p_1}$ be such that $f_0 + f_1 = f$ and $$ ||f_0||_{L^{p_0}} + ||f_1||_{L^{p_1}} \leq 2 ||f||_{L^{p_0} + L^{p_1}}. $$ There must exist such a pair, since either $f=0$ and we can use $f_0 = f_1 = 0$ (here I mean in the a.e. sense) or $||f||_{L^{p_0}+L^{p_1}} > 0$ and we can use the fact that there must exist pairs of functions $(f_0,f_1)$ which come arbitrarily close to realizing the infimum which defines $||f||_{L^{p_0}+L^{p_1}}$. (Note that the 2 can be replaced with $1+\epsilon$ for any $\epsilon > 0$ but I prefer to keep it simple).

Now since $f = f_0 + f_1$ it holds $$||f||_{L^\Psi} = ||f_0 + f_1||_{L^\Psi} \leq ||f_0||_{L^\Psi} + ||f_1||_{L^\Psi}$$ where the last inequality is just the sub-additivity of the norm applied to $||\cdot||_{L^\Psi}$.

Note that if we can show both \begin{align} ||f_0||_{L^\Psi} &\leq ||f_0||_{L^{p_0}} \\ ||f_1||_{L^\Psi} &\leq ||f_1||_{L^{p_1}} \end{align} Then we will have the chain \begin{align} ||f||_{L^\Psi} &\leq ||f_0||_{L^\Psi} + ||f_1||_{L^\Psi} \\ &\leq ||f_0||_{L^{p_0}} + ||f_1||_{L^{p_1}} \\ &\leq 2 ||f||_{L^{p_0} + L^{p_1}} \end{align} which will prove the bound with $B = 2$.

Now that we have the plan laid out, let's get to the technical part. First note that the definition of $||\cdot||_{L^\Psi}$ implies $$ \int_X \Psi\left ( \frac{|f_0(x)|}{||f_0||_{L^{p_0}}} \right ) d\mu \leq 1 \implies ||f_0||_{L^\Psi} \leq ||f_0||_{L^{p_0}} $$ since if the left holds, then $||f_0||_{L^{p_0}}$ is in the set over which the infimum is taken in the definition of $||f_0||_{L^\Psi}$. An analogous fact holds for $f_1$ and $||f_1||_{L^{p_1}}$.

Let's prove the first bound \begin{align} \int_X \Psi\left ( \frac{|f_0(x)|}{||f_0||_{L^{p_0}}} \right ) d\mu &= \int_X \left [ \int_0^{|f_0(x)|/||f_0||_{L^{p_0}}} \psi(u)du\right ] d\mu \\ &\leq \int_X \left [ \int_0^{|f_0(x)|/||f_0||_{L^{p_0}}} u^{p_0-1}du\right ] d\mu & (*)\\ &= \int_X \frac{1}{p_0} \frac{|f_0(x)|^{p_0}}{||f_0||_{L^{p_0}}^{p_0}} d\mu \\ &= \frac{1}{p_0}\int_X \left | \frac{|f_0(x)|}{||f_0||_{L^{p_0}}} \right |^{p_0}d\mu \\ &= \frac{1}{p_0} \end{align} In this case using the fact that $p_0 \geq 1$ it follows \begin{align} \int_X \Psi\left ( \frac{|f_0(x)|}{||f_0||_{L^{p_0}}} \right ) d\mu \leq \frac{1}{p_0} \leq 1 \implies ||f_0||_{L^{\Psi}} & \leq ||f_0||_{L^{p_0}} \end{align} which shows the first inequality we need. The line $(*)$ uses the fact that if $u \in [0,1]$ then $\psi(u) = u^{p_1 - 1} \leq u^{p_0 - 1}$ since $1 \leq p_0 \leq p_1$. The last equality uses the fact

The next inequality will use a similar approach \begin{align} \int_X \Psi\left ( \frac{|f_1(x)|}{||f_1||_{L^{p_1}}} \right ) d\mu &= \int_X \left [ \int_0^{|f_1(x)|/||f_1||_{L^{p_1}}} \psi(u)du\right ] d\mu \\ &\leq \int_X \left [ \int_0^{|f_1(x)|/||f_1||_{L^{p_1}}} u^{p_1-1}du\right ] d\mu & (**)\\ &= \int_X \frac{1}{p_1} \frac{|f_1(x)|^{p_1}}{||f_1||_{L^{p_1}}^{p_1}} d\mu \\ &= \frac{1}{p_1}\int_X \left | \frac{|f_1(x)|}{||f_1||_{L^{p_1}}} \right |^{p_1}d\mu \\ &= \frac{1}{p_1} \end{align} In this case using the fact that $p_1 \geq 1$ it follows \begin{align} \int_X \Psi\left ( \frac{|f_1(x)|}{||f_1||_{L^{p_1}}} \right ) d\mu \leq \frac{1}{p_1} \leq 1 \implies ||f_1||_{L^{\Psi}} & \leq ||f_1||_{L^{p_1}} \end{align} which shows the first inequality we need. The line $(**)$ uses the fact that if $u \geq 1$ then $\psi(u) = u^{p_0 - 1} \leq u^{p_1 - 1}$ since $1 \leq p_0 \leq p_1$.

This completes the proof using $B = 2$ which can be improved to $B = 1 + \epsilon$ in essentially the same way.

To see why $1 \leq p_0, p_1$ note that if this isn't the case then $\Psi$ isn't convex. To see why $p_0 \leq p_1$, suppose that the opposite holds. Then $L^{p_1} \subset L^{p_0}$ and if $f \in L^{p_0} \setminus L^{p_1}$ it always holds $||f||_{L^{p_0} + L^{p_1}} < \infty$ but it may be $||f||_{L^{\Psi}}$ (you should check this for yourself to reinforce your understanding)!

Answer 2

I came up with an alternative answer after a night's sleep. But frankly, I like Matt Werenski's answer a lot more. That being said, I am happy to share my own answer as it uses a rather different approach. We also assume $1 \leq p_0 < p_1$.

By definition of the norm in $L^{p_0} + L^{p_0}$, there is a decomposition of $f$ as $f = f_0 + f_1$ with $f_0 \in L^{p_0}$ and $f_1 \in L^{p_1}$ and $\|f_0\|_{L^{p_0}} + \|f_1\|_{L^{p_1}} = \|f\|_{L^{p_0}+L^{p_1}} (1 + \epsilon)$. (I will omit the $\epsilon$ from now on as we will let it tend to 0 in the end any way.) Without loss of generality, we can also assume $f, f_0, f_1$ are all non-negative.

For any $M>0$, we can further decompose $f_1$ as $f_1=f_1' + f_1''$ $$ \begin{align} f_1'(x) &= \min \left(M, f_1(x)\right) \\ f_1''(x) &= f_1(x) - f_1'(x) \leq f_1(x) \chi_{f_1(x)>M}. \end{align} $$

It is easy to show $f_1''$ is also in $L^{p_0}$ and $\|f_1''\|_{L^{p_0}} \leq \|f_1''\|_{L^{p_1}} \leq \|f_1\|_{L^{p_1}}$. Our intention is to add $f_1''$ to $f_0$ eventually.

Similarly, we can decompose $f_0 \in L^{p_0}$ as $f_0=f_0' + f_0''$ where $$ \begin{align} f_0'(x) &= \min(f_0(x),M - f_1'(x)) \leq M \\ f_0''(x) &= f_0(x) - f_0'(x) \ge 0. \end{align} $$

Notice that $f_0'$ is also in $L^{p_1}$ and $\|f_0'\|_{L^{p_1}} \leq \|f_0'\|_{L^{p_0}} \leq \|f_0\|_{L^{p_0}}$.

This allows us to define $F_0 \in L^{p_0}$ and $F_1 \in L^{p_1}$ $$ \begin{align} F_0 &= f_0'' + f_1'' \\ F_1 &= f_0' + f_1'. \end{align} $$

with $\|F_0\|_{L^{p_0}} + \|F_1\|_{L^{p_1}} \leq \|f_0''\|_{L^{p_0}} + \|f_1''\|_{L^{p_0}} + \|f_0'\|_{L^{p_1}} + \|f_1'\|_{L^{p_1}} \leq 2 \left( \|f_0\|_{L^{p_0}} + \|f_1\|_{L^{p_1}}\right)$.

It is easy to see that $F_1(x) = \min(M, f(x))$. What is special about the new decomposition $f = F_0 + F_1$ is that now $\int_X \Psi(f/M) d\mu$ can be written as the sum of two terms, of which the first only depends on $F_0$ and the second only on $F_1$. After some calculations we see the claim is satisfied with $B=2$.