Equivalence of weak $L^p$ norms

Question

I'm kind of new to the subject of weak $L^p$ spaces. The definition of the (quasi-)norm in weak $L^p$ ($p\in(0; \infty)\,$) over $\sigma$-finite measure space $(X, \mu)$ I use is $||f||_{L^{p, \infty}} = \sup_{t\in\left(0, \infty\right)} t^{\frac{1}{p}}f^*\left(t\right)$, where $f^*(t)=\inf\{\lambda>0; \mu_f(\lambda)\leq t\}$ and $\mu_f(\lambda) = \mu(\{x\in X; |f(x)|>\lambda\})$. However, when studying another article regarding weak $L^p$ spaces, I came across this definition $||f||_{L^{p, \infty}} = \sup_{\lambda\in\left(0, \infty\right)} \lambda (\mu_f(\lambda))^\frac{1}{p}$.

Somehow I forced myself to believe $\sup_{t\in\left(0, \infty\right)} t^{\frac{1}{p}}f^*\left(t\right) = \sup_{\lambda\in\left(0, \infty\right)}\lambda(\mu_f\left(\lambda\right))^\frac{1}{p}$. Today I decided to prove it rigorously. To my surprise, I've been struggling to prove it. I proved that $f^*(\mu_f(\lambda))\leq\lambda$ (if $\mu_f(\lambda)<\infty$) and $\mu_f(f^*(t))\leq t$ (if $f^*(t) < \infty)$ but I failed in using it to prove the equality.

Does the equality really hold? I believe I must be missing some simple thing(s) but after spending hours trying to prove it I find myself really desperate.

Thank you for any help!

Answer 1

I will first prove that $$ t^{\frac{1}{p}}f^*\left(t\right) \leq \sup_{\lambda\in\left(0, \infty\right)}\lambda(\mu_f\left(\lambda\right))^\frac{1}{p} + \epsilon $$ holds for every $t\geq 0$ and small $\epsilon>0$.

Lets fix $t\geq 0$. If $f^*(t)=0$ we are done. Now, assume it is positive.

Also define $\lambda_1$ such that $0<\frac{f^*(t)}{2}<\lambda_1<f^*(t)$ and $f^*(t)-\lambda_1<\frac{\epsilon}{t^{1/p}}$. We can do this if $\epsilon$ is suff. small. By its definition we have $\mu_f(\lambda_1)>t$ and $\lambda_1 + \frac{\epsilon}{t^{1/p}} > f^*(t)$. Therefore we have \begin{align*} t^{\frac{1}{p}}f^*\left(t\right) & \leq t^{1/p}(\lambda_1 + \epsilon)\\ & = t^{1/p}\lambda_1 + t^{1/p}\frac{\epsilon}{t^{1/p}}\\ & \leq (\mu_f(\lambda_1))^{1/p}(\lambda_1) + \epsilon\\ & \leq \sup_{\lambda\in\left(0, \infty\right)}\lambda(\mu_f\left(\lambda\right))^\frac{1}{p} + \epsilon \end{align*}

Now lets show

$$ \sup_{t\in\left(0, \infty\right)}t^{\frac{1}{p}}f^*\left(t\right) \geq \lambda(\mu_f\left(\lambda\right))^\frac{1}{p} - \epsilon_* $$ holds for every $\lambda > 0$ and small $\epsilon_*>0$.

Lets again take a fixed $\lambda_*>0$. Define $t_*=\mu_f(\lambda_*)$. If $t_*=0$ we are done. Assume $t_*>0$. Define $\lambda_0$ such that $$ \lambda_0 = \inf \{ \lambda>0 : \mu_f(\lambda) < \mu_f(\lambda_*) - \epsilon \} $$ where $\epsilon = (\frac{\epsilon_*}{\lambda_*})^p$. Since $\mu_f(\lambda)$ is non increasing, we should have $\lambda_0 > \lambda_*$. If there is no such $\lambda_0$ then we have $\mu_f(\lambda)\geq\mu_f(\lambda_*) - \epsilon$ for all $\lambda > \lambda_*$. Taking $\lambda \rightarrow \infty$ we have $\mu_f(\lambda_*)=0$ which contradicts with $t_*>0$. Therefore such $\lambda_0$ exists. By its definition we also have $f^*(t_*-\epsilon)=\lambda_0>\lambda_*$. Therefore we have

\begin{align*} \lambda_* t_*^{1/p} - \epsilon_* & \leq \lambda_* t_*^{1/p} - \lambda_* \epsilon^{1/p}\\ & \leq \lambda_* (t_*^{1/p} - \epsilon^{1/p})\\ & \leq \lambda_* (t_* -\epsilon)^{1/p}\\ & \leq f^*(t_*-\epsilon) \ (t_*-\epsilon)^{1/p}\\ & \leq \sup_{t\in\left(0, \infty\right)}t^{\frac{1}{p}}f^*\left(t\right) \end{align*} Here, from second line to third line, when $\epsilon$ is suff. small, it should work.

Edit: Justification for the last sentence.

• For $p=1$ it is true.
• When $p>1$, consider the function $\phi(\epsilon) = (t_* -\epsilon)^{1/p} + \epsilon^{1/p}$. The derivative $\phi'(\epsilon) = -\frac{1}{p}(t_* -\epsilon)^{1/p-1} + \frac1{p}\epsilon^{1/p-1}>0$ for small $\epsilon>0$. Therefore $t^{1/p}-\epsilon^{1/p} \leq (t_* -\epsilon)^{1/p}$
• If $p<1$ we need to slightly modify the definition of $\epsilon.$ Define it as $\epsilon = (\frac{\epsilon_*}{\lambda_*})^{1/p}$. Then function $\phi$ becomes $\phi(\epsilon) = (t_* -\epsilon)^{1/p} + \epsilon^{p}$. Now $\phi'(\epsilon) = -\frac{1}{p}(t_* -\epsilon)^{1/p-1} + p\epsilon^{p-1}>0$ for small $\epsilon>0$. Hence $t^{1/p}-\epsilon^{p} \leq (t_* -\epsilon)^{1/p}$

Answer 2

Here is what I consider a less technical proof: Lets start proving $$ \sup_{t>0}t^{\frac 1p}f^*(t) \le \sup_{\lambda>0}\lambda \mu_f(\lambda)^{\frac 1p}. $$

Fix $t>0$. If $\mu_f(\lambda)>t$ for every $\lambda>0$, then $\mu(\{|f|=\infty\})\ge t$ and both sides would be $\infty$, so we can safely assume there is a $\lambda>0$ such that $\mu_f(\lambda)\le t$. If we suppose that $\mu_f(\lambda')\le t$ for every $\lambda'>0$, then $f^*(t)=0$, and $t^{\frac1p}f^*(t)=0\le \sup_{\lambda>0}\lambda \mu_f(\lambda)^{\frac 1p}$. Therefore, we can also safely assume that there exists $\lambda'$ such that $\mu_f(\lambda)>t$. It should not be hard to convince yourself that then, since $\mu_f$ is decreasing and right-continuous, there exists a $\lambda$ such that $$ \mu_f(\lambda) \le t \le \lim_{\beta\uparrow\lambda} \mu_f(\lambda). $$ Therefore, since $f^*$ is decreasing, using one of your inequalities $$ t^{\frac1p}f^*(t) \le \lim_{\beta\uparrow\lambda} \mu_f^{\frac 1p}(\beta)f^*(\mu_f(\lambda)) \le \lim_{\beta\uparrow\lambda} \mu_f^{\frac 1p}(\beta)\lambda = \lim_{\beta\uparrow\lambda} \mu_f^{\frac 1p}(\beta)\beta \le \sup_{\beta>0}\beta\mu_f^{\frac 1p}(\beta), $$ so combining both cases of $t$ and taking supremums, we get the desired inequality.

To check $$ \sup_{\lambda>0}\lambda \mu_f(\lambda)^{\frac 1p} \le \sup_{t>0}t^{\frac 1p}f^*(t), $$ the proof is essentialy the same, but for the $\infty$ case you may need the identity $\mu_{f^*}(\lambda)=\mu_f(\lambda)$. Then, repeating the arguments mutatis mutandis, the proof is complete.