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 Jun4 accepted How can I simplify a function? Jun4 asked How can I simplify a function? Apr3 comment About the property of Littlewood-Paley partition of unity. Thank you for the comment! Apr3 accepted About the property of Littlewood-Paley partition of unity. Apr3 asked About the property of Littlewood-Paley partition of unity. Dec14 comment Question from Evans' PDE book @WillieWong @ Frank So I think we need some help from Wille Wong. Dec14 revised Question from Evans' PDE book added 19 characters in body Dec14 answered Question from Evans' PDE book Dec13 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 $\mu$. Dec13 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. Dec13 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 $H_0^1$. Dec11 awarded Teacher Dec11 awarded Editor Dec11 revised Question on proof in Evans PDE added 76 characters in body Dec11 answered Question on proof in Evans PDE Dec6 awarded Supporter Dec5 awarded Scholar Dec5 accepted An upper bound for an integral. Dec5 comment An upper bound for an integral. Thank you very much! Dec5 awarded Student