Is it true that for some real $K>0$ and all real $u\in C_0^\infty((0,1))$ we have $$\int_0^1 (u'(x)^2+u(x)^2)\,dx\,\int_0^1 u(x)^2\,dx \le K\Big(\int_0^1 x\,u'(x)^2\,dx\Big)^2\text{ ?}$$
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asked Nov 6, 2023 at 12:52
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No. If $u$ contains a bump of height $1$ and small width $h$, then $\int u'^2\gtrsim 1/h$. On the other hand, this part of $u$ only contributes $\lesssim 1$ to $\int xu'^2$ if we move it close to $x=0$. We can make $\int u^2\simeq 1$ by making $u\simeq 1$ in some other part of the interval, without introducing large derivatives there, so $\int xu'^2$ won't get much larger.
What is true along these lines is $\int u^2\lesssim \int x^2u'^2$; see the answer to Caffarelli-Kohn-Nirenberg-type inequality with nonradial weight. Incidentally, if you tried to go from here to your inequality, you would need Cauchy–Schwarz in reverse.
answered Nov 6, 2023 at 14:49
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$\begingroup$ Can you be more explicit how to construct a function with an arbitrarily high K? I'm struggling to see how you can make $\int u^2$ large enough away from 0, without affecting $\int x u'^2$ in $C^{\infty}_0$. $\endgroup$ Commented Nov 7, 2023 at 12:06
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$\begingroup$ @Eigentime: $u(x)=x/h$ on $0\le x\le h$, then $u(x)=1$ for $h\le x\le 1/2$, and finally $u(x)=2-2x$ on $1/2\le x\le 1$. This is not in $C_0^{\infty}$, but you can slightly modify it (smooth out the corners, shift it slightly to the right), or observe that the inequality, if it were true for $u\in C_0^{\infty}$, would also hold for this $u$ by approximation. $\endgroup$ Commented Nov 7, 2023 at 14:01
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