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For $ v_i \in L^\infty \cap L^2 $, has a compact support, $v_i : \mathbb R^n \to \mathbb R$, $$ \| v_1 \cdots v_k \|_2 \leqslant \|v_k \|_2 \|v_1 \|_\infty \cdots \| v_{k-1} \|_\infty$$ holds? Then why?

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Please avoid titles that are entirely in $\LaTeX$. – J. M. Jul 22 '12 at 15:26

2 Answers 2

up vote 2 down vote accepted

By definition, $v_i\leq\|v_i\|_\infty=\operatorname{ess\,sup}\limits_{x\in\mathbb{R}^n}\;|v(x)|$ almost everywhere . Thus $$\int (v_1v_2\cdots v_k)^2\,dm\leq\int\|v_1\|_\infty^2\|v_2\|_\infty^2\cdots\|v_{k-1}\|_\infty^2 v_k^2\,dm=\|v_1\|_\infty^2\|v_2\|_\infty^2\cdots\|v_{k-1}\|_\infty^2\int v_k^2\,dm$$ and then taking the square root of both sides we arrive at $$\|v_1\cdots v_k\|_2\leq\|v_1\|_\infty\|v_2\|_\infty\cdots\|v_{k-1}\|_\infty\|v_k\|_2$$

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Oh thank you, it was easy.. – Bamily Jul 22 '12 at 15:38

Write almost everywhere $$|v_1\dots v_k|=\color{red}{|v_1\dots v_{k-1}|}|v_k|\leq \color{red}{\prod_{j=1}^{n-1}\lVert v_j\rVert_{\infty}}|v_k|.$$ We conclude taking the square and integrating.

Note that the fact that the functions are compactly supported is not necessary.

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Thank you Daivde. – Bamily Jul 22 '12 at 15:41

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