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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 $2<p<\infty$ $$\|f\|_p\le C^{1-\frac2p}(\sum\|f_j\|_p^p)^\frac1p\ \ \ ?$$

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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 $$ \|f\|_4^4 = 32/3 > 8 = \|f_1\|_4^4 + \|f_2\|_4^4$$

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