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I'm trying to prove the following: If $L$ is a uniformly elliptic PDE, without costant or linear terms and with essentialy bounded coefficents $a^{ij}$. If $u\in H^1(B_2)$ satisfies $Lu = f$, then $$||u||_{L^{\infty}(B_1)} \le C(||u||_{L^2(B_2)} + ||f||_{L^{\frac{n}{2}}(B_2)}),$$ at least as long $||f||_{L^{\frac{n}{2}}}$ is small enough.

I have seen a similar result for the Laplacian and I believe it should work but I'm having trouble:

The proof for the case of the Laplacian uses estimates that come from the inequality $$\int\phi^2u^{p - 2}|\nabla u|^2 \le (\dfrac{2}{p - 1})^2 \int|\nabla \phi|^2 u^p - \dfrac{2}{p - 1}\int\phi^2u^{p - 1}\Delta u,$$

where $\phi$ is some cut off function. This inequality uses explicitly the laplacian and is proved via the divergence theorem, but in our case $u\in H^1$ so the Laplacian is not defined. I do not see how to change this inequality in some way that $L$ apears instead of $\nabla$, and without $L$ I do not know how to make $f$ appear suddenly.

In the cases I've seen for the Laplacian what they actually prove is that $$||u||_{L^{\infty}(B_1)} \le C(||u||_{L^2(B_2)}) $$ if $||f||_{L^{\frac{n}{2}}(B_2)}$ is small, so $f$ does not appear in the bounds. Can the inequality that I'm trying to prove be improved to an inequality of this form or there is a reason why $f$ has to appear in the right hand side?

If the inequality with the Laplacian can't be fixed is there any suggestion on how to make $L$ actually appear. Sobolev/Poincare inequalities seem to only make the gradient appear, not L (hence f), so I do not really know what to do.

Thanks!

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You want to apply the standard bootstrapping getting $L^q$ for all $q$, followed by "Moser iterations" i.e. tracking the constants to push for $L^\infty$. I will sketch the argument in the simpler case of the Laplacian and where $u$ is zero on the boundary, leaving the details for you as an exercise.

Supposing first of all that $u\in H^1_0(B_2)$. Then $$ - \Delta u = f $$ implies, multiplying by $u$ and itegrating by parts $$\int_{B_2}|\nabla u|^2 dx =\int_{B_2} -\Delta u \cdot udx = \int_{B_2}f\cdot u dx \leq \frac{1}{2}(\int_{B_2} f^2 dx +\int_{B_2} u^2 dx ) $$ where we applied the Cauchy-Schwartz inequality. This is sometimes referred to as the Caccioppoli inequality. The key point is that it gives you higher integrability for $u$ when combined with the Sobolev inequality.

Going further, we can multiply the equation by $u^\alpha$ to get $$ \int_{B_2} |\nabla (u^{\frac{\alpha+1}{2}})|^2dx \leq\frac{C}{\alpha+1}(\int_{B_2} f \cdot u^\alpha dx) \leq \frac{C}{\alpha+1} \left(\left(\int_{B_2} |u|^{\alpha \frac{n}{n-2}} dx\right)^\frac{2n}{n-2} + \left( \int_{B_2} |f|^{\frac{n}{2}} dx \right)^\frac{4}{n} \right) $$ where we applied Holder to arrive at the crucial exponenet $\frac{n}{2}$. This then gives $$\left( \int_{B_2} |u|^{(\alpha +1)\frac{n}{n-2} } \right)^\frac{n-2}{n} \leq \frac{C}{\alpha+1} \left(\left(\int_{B_2} |u|^{\alpha \frac{n}{n-2}} dx\right)^\frac{2n}{n-2} + \left( \int_{B_2} |f|^{\frac{n}{2}} dx \right)^\frac{4}{n} \right) $$ by applying the Sobolev inequality. This is a more general Caccioppoli inequality which holds for all $\alpha\geq1$, and applying it iteratively we arrive at $$\| u \|_{L^q(B_2)} \leq C(q) \left(\|u\|_{L^2(B_2)} + \|f\|_{L^{\frac{n}{2}}(B_2)}\right)^{R(q)} $$ for some $C(q)$ and exponent $R(q)$. Now it is for you to carefully go through this argument to estimate $C(q)$ and $R(q)$ explicitly and show that they are in fact uniformly bounded in $q$, hence we can pass to this limit as $q \to \infty$ and deduce $$ \| u\|_{L^\infty(B_2)} \leq C\left( \|u\|_{L^2(B_2)} + \|f\|_{L^{\frac{n}{2}}(B_2)} \right) $$.

Once you have followed this argument, the next step is when $u\in H^1(B_2)$. You then have to introduce a test function $\chi$ which satisfies

$\chi\equiv 1 $ on $B_{R''}$, $\chi\equiv 0$ outside $B_{R''}$ and $|\nabla \chi| \leq \frac{C}{R'-R''}$

to cut off at the boundary. Now multiply through the equation instead by $u^\alpha \chi^2$. This is indeed a bit messy, but dealing with the lower order terms we eventually get a general Caccioppoli type inequality of the form $$ \left( \int_{B''} |u|^{(\alpha +1)\frac{n}{n-2} } \right)^\frac{n-2}{n} \leq \frac{C}{(\alpha+1)(R'-R'')^2} \left(\left(\int_{B_{R'}} |u|^{\alpha \frac{n}{n-2}} dx\right)^\frac{2n}{n-2} + \left( \int_{B_{R'}} |f|^{\frac{n}{2}} dx \right)^\frac{4}{n} \right) $$

Then choosing a sequence of radii $R_2 > R^{\alpha_k} > R_1$ so that $R^{\alpha_k}$ is a decreasing sequence of radii, indexed by $\alpha_k \to \infty$, which tends to $R_1$ as $k\to \infty$. Applying the Caccioppoli inequality iteratively on this sequence of discs, we get $L^q(B_1)$ for all $q$, and track the constants to show that you can push to the limit as $q\to \infty$ and so get $L^\infty(B_1)$.

Once you have followed this argument, run the whole thing again for your uniformly elliptic operator. Everything goes the same as for the Laplacian except you will have to carry around your constants of ellipticity.

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  • $\begingroup$ I should probably point out, in case I was too brief in sketching these arguments, that you should be able to find this type of calculation done thoroughly in Evans, Gilbarg & Truidinger or any other standard introductory text for elliptic PDE :) $\endgroup$ – Aerinmund Fagelson Oct 2 '16 at 16:42
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For constant coefficents, I recalled you can rotate the coordinate to get it into the standard form, then apply your result.

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