Recall given any function $v(x)$ defined on $B$ (the unit ball centred at the origin in $ R^N$) we can write 


$$v(x) = \sum\_{k=0}^\infty a\_k(r) \psi\_k(\theta)$$ 


where $ r=|x|$ and $ \theta = \frac{x}{|x|}$ and $ \psi\_k$ are the $L^2(S^{N-1})$ normalized eigenfunctions. 


We now suppose $u(x)$ satisfies 


$$ -\Delta u(x) = f(x)$$ in $ B$ with $u=0$ on $ \partial B$. There is the standard maximum principle estimate that 
$$ \sup\_B |u| \le \frac{1}{2N} \sup\_B |f|.$$ 


QUESTION If we assume that both $u$ and $f$ have no $k=0$ modes can we obtain a better maximum principle estimate? 


any comments would be greatly appreciated. 



Improved maximum principle estimates (deleting first mode)
Yes, one can get a constant better than $1/(2N)$. Simply take a look at the proof of the original bound $\|u\|\_\infty \le \|f\|\_\infty / (2N)$: one writes
$$ u(x) = \int\_B G\_B(x, y) f(y) dy , $$
estimates $|f(y)|$ by $\|f\|\_\infty$ under the integral, and uses $$\int\_B G\_B(x, y) dy = (1 - |x|^2) / (2 N) \le \frac{1}{2 N}.$$ If we know in advance that $f$ (and hence also $u$) is orthogonal to the first eigenfunction $\phi\_0$, then for any $\alpha$ we can write
$$ u(x) = \int\_B (G\_B(x, y) - \alpha \phi\_0(y)) f(y) dy $$
and again estimate $|f(y)|$ by $\|f\|\_\infty$. This leads to
$$ \|u\|\_\infty \le \|f\|\_\infty \times \sup\_{x \in B} \inf\_{\alpha \in \mathbb{R}} \int\_B |G\_B(x, y) - \alpha \phi\_0(y)| dy . $$
Noticing that for $\alpha > 0$ small enough we have
$$ \int\_B |G\_B(x, y) - \alpha \phi\_0(y)| dy < \int\_B G\_B(x, y) dy = \frac{1 - |x|^2}{2 N} \, ,$$
we easily find that $\|u\|\_\infty \le C \|f\|\_\infty$ for some $C < 1 / (2 N)$.


