The seminorm of $L^{p,\infty}$ does not fulfill the triangular inequality.

Question

Hello I am attending a course of harmonic analysis. My professor pointed out a fact about $L^{p,\infty}(\mathbb{R^{d}})$ spaces, leaving the proof to us, and I got stuck. It is not an assignment of any kind, he simply stated it as a curiosity. The fact is the following: given two functions $f,g \in L^{p,\infty}$ then we have $$ \left\Vert f +g \right\Vert_{p,\infty} \leq c_{p} \left[ \, \left\Vert f \right\Vert_{p,\infty}+ \left\Vert g \right\Vert_{p,\infty} \right]$$ where $c_{p}=\max\left\{ 2,2^{\frac{1}{p}}\right\}$ and $$ \left\Vert f \right\Vert_{L^{p,+\infty}}:=\inf \left\{ c : \mathcal{L}(\left\{ \left| f \right|>\lambda \right\}) < \frac{c^{p}}{\lambda^{p}} \forall \lambda>0 \right\}$$ obviously $\mathcal{L}$ is the $d-$dimensional Lebesgue measure. I proceded as follows: Given $\epsilon>0$ $$ \mathcal{L}\left( \{ \left\vert f +g \right\vert > \lambda \} \right) \leq \mathcal{L}(\{ \vert f \vert> \frac{\lambda}{2} \})+\mathcal{L}(\{ \vert g \vert> \frac{\lambda}{2} \}) \\ \leq \frac{2^{p}\left( \Vert f \Vert_{p,\infty} +\epsilon \right)^p}{\lambda^{p}}+\frac{2^{p}\left( \Vert g \Vert_{p,\infty} +\epsilon \right)^p}{\lambda^{p}}.$$ As a consequence, we obtain that for every $\epsilon>0$ $$ \Vert f+g \Vert_{p,\infty} \leq 2 \, \left[ (\Vert f\Vert_{p,\infty} +\epsilon )^{p} + (\Vert g\Vert_{p,\infty} +\epsilon )^{p} \right]^{\frac{1}{p}}.$$ How can I conclude? I am only able to prove the inequality for $c_{p}=2.$ Moreover is the constant $c_{p}$ sharp? Are there any functions for which we have the equality?

Answer 1

When $p\geq1$ you have, for $a,b>0$, $$ (a^p+b^p)^{1/p}\leq a+b, $$ and you are done. When $p<1$, you can use $$ (a^p+b^p)^{1/p}\leq 2^{1/p} (a+b). $$