Leun Kim
Reputation
Top tag
Next privilege 100 Rep.
Edit community wikis
 Jun 4 accepted How can I simplify a function? Jun 4 asked How can I simplify a function? Apr 3 comment About the property of Littlewood-Paley partition of unity. Thank you for the comment! Apr 3 accepted About the property of Littlewood-Paley partition of unity. Apr 3 asked About the property of Littlewood-Paley partition of unity. Dec 14 comment Question from Evans' PDE book @WillieWong @ Frank So I think we need some help from Wille Wong. Dec 14 revised Question from Evans' PDE book added 19 characters in body Dec 14 answered Question from Evans' PDE book Dec 13 comment Question on proof in Evans PDE @Euler....IS_ALIVE So if $C'$ does not depend on $\lambda_i$, then we can say that for all $i=1,2,\cdots$, $\| u_i \|_{H_0^1} \leqslant C'$ uniformly which means that the set $K$ is bounded by $C'$ in $H_0^1$ with sufficiently large $\mu$. Dec 13 comment Question on proof in Evans PDE @Euler....IS_ALIVE If that constant $C'$ depends on $\lambda$, we cannot say the boundedness of the set. Because if $C'$ depends on $\lambda$ and if there are infinitely many $\lambda's$(denote$\lambda_i, i=1,2,\cdots$), then of course for every $\lambda_i$'s, the corresponding $u_i$ satisfies $\| u_i \|_{H_0^1} < \infty$. But $\lim_{i \to \infty} \| u_i \|_{H_0^1}$may diverge. Dec 13 comment Question on proof in Evans PDE @Euler....IS_ALIVE Yes, we assumed $u \in H_0^1$, but we could not know the dependence of $\lambda$. What I wrote means that C' does not depend on $\lambda$ so that the set K:= $\{ u \in H_0^1 | u = \lambda A[u] \; \text{for some} \; 0 \leqslant \lambda \leqslant 1 \}$ is bounded in $H_0^1$. Dec 11 awarded Teacher Dec 11 awarded Editor Dec 11 revised Question on proof in Evans PDE added 76 characters in body Dec 11 answered Question on proof in Evans PDE Dec 6 awarded Supporter Dec 5 awarded Scholar Dec 5 accepted An upper bound for an integral. Dec 5 comment An upper bound for an integral. Thank you very much! Dec 5 awarded Student