A proposition about bounded and weak-$L^2$ integrable fucntion

Question

I have this proposition that I have no idea how to start. any hints helps.

If $f\in L^{2,\infty}\cap L^{\infty}$ then $f\in L^p$ for $p\geq2$

and the second part asks me to generalize this result.

$L^{p,\infty}:=\left\{f:||f||_{p,\infty}:= \text{sup}_{\lambda>0}\lambda \mu(\{|f(x)|>\lambda\})^{1/p} <\infty\right\}$.

Answer 1

This is a partial answer to your question. I haven't managed to solve the case $p=2$ yet.

By Fubini's theorem, we may write $$ \int_\Omega |f|^p\, d\mu = p \int_0^\infty t^{p-1} \mu(\{|f|> t\}) dt. $$ Let $f\in L^{2,\infty}(\Omega)\cap L^{\infty}(\Omega)$ and suppose that $p> 2$. By our assumption, we have $$ \sup_{t\ge 0} t^2 \mu(\{|f|> t\}) = N < \infty. $$

Since $f$ is essentially bounded, the left hand side reduces to $$\begin{align} \int_\Omega |f|^p\, d\mu &= p \int_0^M t^{p-1} \mu(\{|f|> t\}) dt \\ &= p \int_0^M t^{p-3} t^2\mu(\{|f|> t\}) dt \\ &\le pN \int_0^M t^{p-3}\, dt. \end{align}$$

This integral converges iff $p-3>-1$, i.e. $p>2$.

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It is easily seen that this method can be slightly modified into proving that $$ L^{p,\infty}(\Omega) \cap L^\infty(\Omega)\subset L^q(\Omega) $$ whenever $q>p$. As before, the method doesn't work for $q=p$ since $t^{-1}$ is not integrable in any neighborhood of $0$.

Answer 2

BigbearZzz has shown how to deal with the cases $p>2$. The reason that their technique breaks down in the case $p=2$ is that the result is false there. This answer is intended to supplement their answer by giving a counterexample in the case $p=2$.

Consider $(0, \infty)$ equipped with the Lebesgue measure and let $f(x) = x^{-1/2} {\bf 1}_{\{x \geq 1\}}$. Then $f$ is not in $L^2(0, \infty)$ (by comparison to the harmonic series). Clearly $f \in L^\infty(0,\infty)$ with $\|f\|_\infty \leq 1$.

It remains to show that $\|f\|_{2,\infty} < \infty$. First note that since $\|f\|_\infty \leq 1$, we only have to compute $\mu(|f| > t)$ for $t \in (0,1]$. This we can do explicitly. For $t \in (0,1]$ we have,

\begin{align*} \mu(|f| > t) &= \mu(\{ x^{-1/2} > t \} \cap \{ x \geq 1 \}) \\ &= \mu( [1, \frac{1}{t^2})) = \frac{1}{t^2} - 1 \end{align*}

So $t^2 \mu(|f| > t) = 1 - t^2 \leq 1$ for $t \in (0,1]$. In total, this gives us that $\|f\|_{2,\infty} \leq 1 < \infty$. This means that $f \in L^{2,\infty}(0,\infty) \cap L^\infty(0,\infty)$ but $f \not \in L^2(0,\infty)$ and so $f$ is the desired counterexample.