I am confused about the proof of the following theorem which is found in Evans' PDEs, Chapter 6.3.

Theorem 3(the infinite differentiability in the interior). Assume $$ a^{ij},b^i,c\in C^\infty(U),(i,j=1,...,n) $$ and $f\in C^\infty(U)$. Suppose $u\in H^1(U)$ is a weak solution of the ellipctic PDE $$ Lu=f \quad\text{in }U $$ Then $u\in C^\infty(U)$.

Proof: From Theorem 2, we have $u\in H^m_{loc}(U)$ for each $m=1,2,....$. By General sobolev inequality, we have $u\in C^k(U)$ for each $k=1,2,...,$.

I don't understand why we can say "$u\in C^k(U)$ for each $k=1,2,...$". The Sobolev inequality should only give us that $u\in C_{loc}^k(U)$.


1 Answer 1


This is because of the notation choice of Evans, which you can find in Appendix A at the back of the book. Specifically, we have the definition

$$ C^k(U) := \{ u : U \to \mathbb R \mid u \text{ is } k\text{-times continuously differentiable} \} $$ In particular, the function is allowed to explode at the boundary, if $U$ is an open domain. Therefore, a function in $C^k_{\text{loc}}(U) $, under Evans' notation, is a function in $C^k(U)$. Here's a partial list for comparison (again, for the full list see Appendix A)

$C(U)=\{u: U \rightarrow \mathbb{R} | u \text { continuous }\}$

$C(\bar{U})=\{u \in C(U) | u$ is uniformly continuous on bounded subsets of $U\}$

$C^{k}(U)=\{u: U \rightarrow \mathbb{R} | u \text { is } k \text { -times continuously differentiable }\}$

$C^{k}(\bar{U})=\left\{u \in C^{k}(U) \middle| \substack{\displaystyle D^{\alpha} u\text{ is uniformly continuous on} \\ \displaystyle\text{ bounded subsets of } U, \text { for all }|\alpha| \leq k }\right\}$

Thus if $u \in C^{k}(\bar{U}),$ then $D^{\alpha} u$ continuously extends to $\bar{U}$ for each multiindex $\alpha,|\alpha| \leq k$

  • $\begingroup$ But in evans book, the general Sobolev inequality used require that the boundary condition is $C^1$. Does that matter? By definition of $H_{loc}$, for any $V\subset\subset U$, $u\in H^m(V)$ then we get that $u\in C^k(V)$. Does it mean $u\in C^k(U)$? Or you mean, it's enough to show that $u\in C^k(B)$ for any open ball in $U$? $\endgroup$
    – Q-Y
    Apr 29, 2020 at 8:42
  • $\begingroup$ Boundary doesn't matter for proof of Theorem 2 of §6.3. Yes, To apply embedding you need $V$ to have nice boundary. So you can't apply it for any $V\Subset U$, you should pick a nice collection of subsets. Yes, like open balls, because in this section, $U$ is an open (bounded) set, so $u\in C^k(\overline V)$ for all such $V$ implies $u\in C^k(U)$. Also, if what you said was true, i.e. for any (and really any, not just smooth boundary) $V\Subset U$, we had $u\in C^k(V)$, then of course this implies $u\in C^k(U)$, from the above definition @Q-Y $\endgroup$ Apr 29, 2020 at 9:10
  • 1
    $\begingroup$ Thank you so much $\endgroup$
    – Q-Y
    Apr 29, 2020 at 15:17

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