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From the Cauchy-Schwarz inequality, we can prove that $$\lVert w(x)\rVert^2_{L^2_{[0,1]}}=\int_0^1 w(x)^2\, dx \leq \sqrt{\int_0^1 w(x) \,dx}\cdot \sqrt{\int_0^1 w(x)^3\, dx}.$$

Is it possible to prove another inequality with the other direction, that is, $$\int_0^1 w(x)^3\, dx \leq C_1\cdot \left(\int_0^1 w(x)^2 dx\right)^{1/n}\;?$$

Thanks a lot!!!

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We have $$\int_0^1 w(x)^3\, dx \leq \max\{w(x)\} \cdot \left(\int_0^1 w(x)^2 dx\right)\;$$ and with some special boundary condition, we can have $\max\{w(x)\}<=c \cdot \|w\|^{1/2} \cdot \|w_x\|^{1/2}$ or $max{w(x)}<=c \cdot \|w_x\|$, but what I want is whether we can get a constant $C_1$ in the above inequality? Thanks. – Sophia Tang Aug 16 '12 at 0:49
:I already gave you the constant $C$ under certain conditions in my answer. – Mhenni Benghorbal Aug 16 '12 at 6:49

No, because $\int_0^1 w(x)^3\ dx$ can be infinite while $\int_0^1 w(x)^2 \ dx$ is finite. Consider e.g. $w(x) = x^{-1/3}$.

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Yes, you made a point. But could we prove the second inequality under some condition? – Sophia Tang Aug 16 '12 at 0:18
If you want $C_1$ to be a constant, and you allow, for example, all polynomials, then it is impossible, because you can e.g. take a polynomial that uniformly approximates $\min(R, x^{-1/3})$ for arbitrarily large $R$. – Robert Israel Aug 16 '12 at 2:01
Thanks a lot for your answer. – Sophia Tang Aug 16 '12 at 17:44

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