Best constant in Weak-$L^p$-triangle inequality What is the best constant $C_p$ in the "triangle inequality"
$$
\| f + g \|_{p,\infty} \le C_p ( \|f\|_{p,\infty} + \|g\|_{p,\infty})$$
for the weak $L^p$ spaces?
Here, I am mostly interested in the case $p \in [1,\infty)$.
Typical proofs show $C_p \le 2$ and I have an example proving $C_p \ge 2^{1/p}$. Moreover, in the limit case $p = \infty$ we have $C_\infty = 1$, which gives the clue that maybe $C_p = 2^{1/p}$ is correct.
 A: After commuting home, the answer came to me ;)
For $t > 0$ and $\theta \in (0,1)$, we have
$$
\{ |f + g| > t \} \subset \{ |f| \ge \theta t \} \cup \{ |g| \ge (1-\theta) t\}.$$
Thus,
$$\lambda(\{|f+g| > t\} t^p \le \frac{\|f\|_{p,\infty}^p}{\theta^p} + \frac{\|g\|_{p,\infty}^p}{(1-\theta)^p}$$
and taking the supremum over $t > 0$,
we have
$$\|f +g\|_{p,\infty}^p \le \frac{\|f\|_{p,\infty}^p}{\theta^p} + \frac{\|g\|_{p,\infty}^p}{(1-\theta)^p}.$$
Now, we choose
$$
\theta = \frac{\|f\|_{p,\infty}}{\|f\|_{p,\infty} + \|g\|_{p,\infty}}$$
and this leads to
$$\|f +g\|_{p,\infty}^p \le 2 (\|f\|_{p,\infty} + \|g\|_{p,\infty})^p.$$
This shows
$C_p \le 2^{1/p}$.
To see $C_p \ge 2^{1/p}$, we can use
$f(x) = 1/x^{1/p}$ on $\Omega = (0,1)$ and $g(x) = f(1-x)$.
Then, $\|f\|_{p,\infty} = \|g\|_{p,\infty} = 1$.
However, $(f+g)(1/2) = 2^{1+1/p}$ is the minimal value of $f+g$. Hence,
$$\lambda_{f+g}( 2^{1+1/p} ) = 1$$
and thus
$$
\|f+g\|_{p,\infty} \ge 2^{1 + 1/p} = 2^{1/p} ( \|f\|_{p,\infty} + \|g\|_{p,\infty} ).$$
