We know that if $ u \in W^{2,4}( \mathbb{R}^8), $ then $$ \int_{\mathbb{R}^8} \Delta u \cdot u dx = - \int_{\mathbb{R}^8} \left| \nabla u \right|^2 dx \leq 0. $$ I wonder if the following integral $$ \int_{\mathbb{R}^8} ( \Delta u)^3 u dx $$ if it is also nonpositive! In fact, I could not see any Green's integration by parts operation can be easily modify this last inetgral and even if we take $ - \Delta u + u = f $ and write $ u $ as convolution of $ f $ with Bessel kernel, I could not obtain a counter-example.
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It is easy to make $\int f(x)f''(x)^3\, dx$ positive: We only need to have a small interval, of size $\delta>0$, say, on which $f\simeq 1$ and $f'$ increases from $0$ to $1$ at constant rate. This contributes a positive term of order $1/\delta^2$ to the integral.
We can make this example $8$-dimensional by taking $u(x_1,y)=f(x_1)g(y)$ for some fixed $g$.
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$\begingroup$ What you do outside this small interval? $\endgroup$Giorgio Metafune– Giorgio Metafune2025-12-31 16:45:18 +00:00Commented 16 hours ago
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1$\begingroup$ @GiorgioMetafune: Any fixed function that connects everything smoothly will do, since the contributions from this part will not be affected when we send $\delta\to 0$. $\endgroup$Christian Remling– Christian Remling2025-12-31 17:25:15 +00:00Commented 15 hours ago