0
$\begingroup$

Edit: In the paper it says $c_1:= c^k/(k-1)^2$, but that is propably a typo, since it works out with $c_1:= c^{k/(k-1)^2}$. From there on, it is not to tough.

I found a remark in a paper where I couldn't follow the proof, despite the fact, that it looks fairly simple.

Claim: Take a sequence $(\Phi_m)$ with $\Phi_0 >0$ and $0< \Phi_m \leq c^n \Phi^k_{n-1}$ for $m \in \mathbb{N}$; $k>1$ and for a constant $c>0$. Then the assertion

$\limsup_{m \to \infty} \Phi_m^{k^{-m}} \leq c_1 \Phi_0$

holds where $c_1:= c^{k/(k-1)^2}$.

As the proof it is given:

Define the sequence $ \Psi_m:=c_1^{m+1-k^{-1}m} \Phi_m.$ Then using the assumption, one obtains

$0 < \Psi_m \leq \Psi_{m-1}^{k}$.

By iteration we then get

$\Psi_m \leq \Psi_{0}^{k^m}.$

This already shows the claim.

So the last step is clear to me. In opposite, I really have no clue how to derive the first inequality from the assumption. I further don't know where that factor $(k-1)^{-2}$ could be coming from.

Thank all of you in Advance.

The problem is from Jürgen Moser's "A New Proof of de Giorgi's Theorem Concerning the Regularity Problem for Elliptic Differential Equations" (which can be found here: https://onlinelibrary.wiley.com/doi/abs/10.1002/cpa.3160130308). It starts at equation (17).

$\endgroup$
0
$\begingroup$

Regarding the Edit, it has now worked out for the edited constant, assuming it was just a typo.

Proof:

Define the sequence

$\Psi_m:=c_1^{m+1-k^{-1}m} \Phi_m$,

with $c_1:= c^{k/(k-1)^2}$. Using the assumption, we get to

$ 0 < \Psi_m =c_1^{m+1-k^{-1}m} \Phi_m \leq c_1^{m+1-k^{-1}m} c^m \Phi_{m-1}^k = c^{\frac{km+k-m}{(k-1)^2}+m} \Phi_{m-1}^k =(c_1^{m-k^{-1}(m-1)}\Phi_{m-1})^k = \Psi_{m-1}^{k}.$

Via Induction we obtain

$\Psi_m \leq \Psi_{0}^{k^m}=(c_1 \Phi_0)^{k^m}.$

That implies

$\Phi_m^{k^{-m}} \leq \Psi_{0} c_1^{-((m+1)k^{-m}+-k^{-m-1}m)} =c_1 \Phi_0 c_1^{-((m+1)k^{-m}+-k^{-m-1}m)} .$

Since $k>1$, the following holds

$ \lim_{m\to \infty} c_1^{-\frac{(m+1)}{k^{m}}+\frac{m}{k^{m+1}}}=1$

This finally yields

$\limsup_{m \to \infty} \Phi_{m}^{1/{\kappa^{m}}} \leq c_1 \Phi_0 .$

$\endgroup$

Your Answer

By clicking "Post Your Answer", you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.