-1
$\begingroup$

Let $f=\sum f_j$ be a finite sum. Assume that $$ \|f\|_2\le(\sum\|f_j\|_2^2)^\frac12$$ $$\|f\|_\infty\le C\max_j\|f_j\|_\infty$$ Then can we conclude that for $2<p<\infty$ $$\|f\|_p\le C^{1-\frac2p}(\sum\|f_j\|_p^p)^\frac1p\ \ \ ?$$

$\endgroup$
2
$\begingroup$

No. Let's take $C=1$ and two $f_j$'s. Partition your measure space $(X,\Sigma, m)$ into three sets $A_1, A_2, A_3$, all of measure $1/3$: take: $$\eqalign{f_1 = 1, f_2 = 1, f = 2 & \text{ on } A_1\cr f_1 = 1, f_2 = -1, f = 0 & \text{ on } A_2\cr f_1 = 2, f_1 = 0, f = 2 & \text{ on } A_3\cr}$$

Then $\|f\|_2^2 = \|f_1\|_2^2 + \|f_2\|_2^2 = 8/3$, while $\|f\|_\infty = 2 = \max(\|f_1\|_\infty, \|f_2\|_\infty)$. But $$ \|f\|_4^4 = 32/3 > 8 = \|f_1\|_4^4 + \|f_2\|_4^4$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.