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Does the following inequality

$$ \sup_{x\in (0,1)}u^2(x)\leq C_1\int_0^1 x u^2(x)\,\textrm{d}x+C_2\int_0^1 x(u')^2(x)\,\textrm{d}x $$

hold for all $ u\in C^1(0,1)$? If so please give me a proof, and a counterexample if not.


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You mean the closed interval $\[0,1\]$? Then what do you get for constant functions? – WimC Feb 23 '12 at 12:07
@DavideGiraudo,@WimC: Sorry, I missed the square, corrected now. Thanks for your comment. – Y.Z Feb 23 '12 at 12:38
up vote 5 down vote accepted

Consider the family of functions

$$ \phi_\epsilon(x) = 1- x^\epsilon $$

for $\epsilon \in (0,1/4)$. Clearly these functions are in $C^1(0,1)$, and $\sup \phi_\epsilon^2 = 1$ for every $\epsilon$.


$$ \int_0^1 x (1-x^\epsilon)^2 \mathrm{d}x = \int_0^1 x - 2 x^{1+\epsilon} + x^{1+2\epsilon} \mathrm{d}x = \frac12 - \frac{2}{2+\epsilon} + \frac{1}{2+2\epsilon} = \frac{\epsilon^2}{(2\epsilon + 2)(\epsilon + 2)} $$


$$ \int_0^1 x (-\epsilon x^{\epsilon -1})^2 \mathrm{d}x = \epsilon^2 \int_0^1 x^{2\epsilon - 1}\mathrm{d}x = \frac{\epsilon^2}{2\epsilon} $$

Hence we have that the RHS is order $O(\epsilon)$ while the left hand side is order 1 for this family of functions, hence the desired uniform bound cannot be achieved.

Another way to see the obstruction is to take the change of variables $y = \log x$. Then your inequality becomes equivalent to $$\sup_{y\in (-\infty,0)} v^2 \lesssim \int_{-\infty}^0 e^{2y} v(y)^2 \mathrm{d}y + \int_{-\infty}^0 (v'(y))^2 \mathrm{d}y$$ This is clearly false by taking $\psi(y)$ a smooth function such that $\psi |_{y < -5} = 1$ and $\psi |_{y > -3} = 0$ and considering the family $\psi_\lambda(y) = \psi(y/\lambda)$ as $\lambda \to \infty$. (Note that the scaling in $y$ is equivalent to raising to a power in $x$.)

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Willie Wong, you forgot $x$ in integrals, so you were solving another problem – Norbert Feb 23 '12 at 13:24
The question asks about $\int_0^1u^2(x)\;x\;\mathrm{d}x$ and $\int_0^1(u'(x))^2\;x\;\mathrm{d}x$ – robjohn Feb 23 '12 at 13:25
@Norbert and robjohn, oops. I didn't see that $x$. – Willie Wong Feb 23 '12 at 14:07
@robjohn and Norbert: fixed. Now there's a counterexample because of the weight. – Willie Wong Feb 23 '12 at 15:06
Nice counterexample (+1) – robjohn Feb 23 '12 at 15:27

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