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In my lecture it was claimed that $L^r\subset L^p+L^q$ if $p<r<q\leq\infty$. This has been proven here for the case $q<\infty$. What about the case $q=\infty$? Is the claim still true? For example, is $L^2\subset L^1+ L^\infty$?

Filippo
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    That proofs includes $q = \infty,$ no? – William M. Feb 11 '22 at 16:46
  • @WilliamM. No, strictly speaking the proof there has an integral with exponent $q$ (called $p_2$ there). But the decomposition shown also works in the case $q=\infty$, even though it is not checked there. – GEdgar Feb 11 '22 at 16:55
  • @GEdgar as I read it, the proof there showed $\mathscr{L}^r \subset \mathscr{L}^p + \mathscr{L}^q \cap \mathscr{L}^\infty$ for all $p < r < q \leq \infty.$ – William M. Feb 11 '22 at 16:58
  • @WilliamM. The inequality $\int_{D_2}|f_2|^{p_2}\leq \int_{D_2}|f_2|^{p}$ only makes sense if $p_2<\infty$, doesn't it? – Filippo Feb 11 '22 at 17:42
  • @WilliamM. Of course, the goal of the calculation was to show that $f_1\in L^{p_1}\land f_2\in L^{p_2}$ and if $p_2=\infty$, then $f_2\in L^{p_2}$ is trivial. So this solves my problem. – Filippo Feb 11 '22 at 17:47
  • I don't know what you think $\mathscr{L}^\infty$ is but they mean bounded functions. So,$|f_2| \leq 1$ implies $f_2 \in \mathscr{L}^\infty.$ – William M. Feb 11 '22 at 17:48
  • @WilliamM. Exactly, that's why I said that proving $f_2\in L^{p_2}$ is trivial if $p_2=\infty$. But I thought that the proof of $f_1\in L^{p_1}$ doesn't work because I got confused with the inequalities. However, your comment prompted me to look at the proof more carefully and I was able to solve my confusion. So thank you for your help! – Filippo Feb 11 '22 at 17:59

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I just show $L^2 \subset L^1 + L^\infty$.

$$ f = f \, {\bf 1_{(|f| \geq 1)}} + f \, {\bf 1_{(|f| < 1)}}$$

The last term is bounded by 1 hence in $L^\infty$. Since ${\bf 1_{(|f| \geq 1)}}$ is less than $|f|$, we get the bound

$$\int |f| \, {\bf 1_{(|f| \geq 1)}} \leq \int |f|^2$$

and thus the first term is in $L^1$.

coudy
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