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I have two problems here.

1) I need to show $\|xf(x)\|_2 \le \|f(x)\|_3$ given $f \in L^3[0,1]$.

My approach: I know $\|f(x)\|_2 \le \|f(x)\|_3$, and hope to show $\|xf(x)\|_2 \le \|f(x)\|_2$. That's true because $x \in [0,1]$, $xf(x) \le f(x)$. Am I right?

2) If $f \in L^{5}(E)$, $E=[0,1]$ and $\int_{E} f(x)dx=0$, then $\int_{E} |1-f|^5 dx \ge 1$.

I think the key is to define $g=1-f$ and use $\|g\|_p \ge \|g\|_2$ for $p \ge 2$. Please give me any confirmation or tell me wrong.

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For the first question you have to put $||$ since e don't know if $f$ is non-negative, but you got the idea. For the second question, you can use the fact that $\lVert f\rVert_5\geq \lVert f\rVert_2$. I don't see how you can conclude with only the norm $1$. – Davide Giraudo Dec 7 '11 at 19:53
thanks. I was thinking of 2 but wrote 1. – Alex J. Dec 7 '11 at 19:58
Your notation is ugly. One takes the norm of functions, not of values. – Jonas Teuwen Dec 7 '11 at 21:17
In $L^p$ space, you can't specify the value of a function at a point of measure $0$. – Jack Dec 7 '11 at 21:29
I would kindly recommend you to avoid use of variables in expressions that are independent of the variables (e.g. $\|f(x)\|$). – AD. Dec 8 '11 at 7:20
up vote 5 down vote accepted

For (1), we can apply Hölder's inequality: $$\|xf\|_{L^2[0,1]}=\sqrt{\int_{[0,1]}x^2f^2}\leq\sqrt{\Big(\int_{[0,1]}x^{2\cdot3}\Big)^{\frac{1}{3}}\Big(\int_{[0,1]}f^{2\cdot\frac{3}{2}}\Big)^{\frac{2}{3}}}=\Big(\int_{[0,1]}x^{6}\Big)^{\frac{1}{6}}\|f\|_{L^3[0,1]}.$$ Now (1) follow from $$\int_{[0,1]}x^{6}=\frac{x^7}{7}\Big|_0^1=\frac{1}{7}\leq 1.$$

For (2), since $\displaystyle\int_{[0,1]}f=0$ by assumption, we have $$1=\int_{[0,1]}1=\int_{[0,1]}(1-f).$$ Now apply the Hölder's inequality to the right hand side, we obtain $$1=\int_{[0,1]}(1-f)\leq \Big(\int_{[0,1]}|1-f|^5\Big)^{\frac{1}{5}}\Big(\int_{[0,1]}1\Big)^{\frac{4}{5}}=\Big(\int_{[0,1]}|1-f|^5\Big)^{\frac{1}{5}},$$ which implies that $$1\leq \int_{[0,1]}|1-f|^5,$$ as required.

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