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seen Nov 15 at 14:10

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