Improved maximum principle estimates (deleting first mode)

Question

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.

Answer 1

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 $C < 1 / (2 N)$.