Let $U$ an open, bounded and convex set in $R^n$. Let $(u_k)_{k \in N}$ a sequence of functions defined in $\overline{U}$. Suposse that each $u_k \in C(\overline{U}) \cap C^{2}(U)$ and harmonic in $U$. Suppose too $0 \leq u_{k+1}(x) \leq u_{k}(x) \leq 1$ for all $x \in \overline{U}$ and for all $k$.

I know how to prove that the function $u(x) = \lim u_k(x), x \in \overline{U}$ is in $C^{2}(U)$ and I know how to prove that $\Delta u = 0$ in $U$ . My doubt is : $u \in C(\overline{U})$ ? I tried to prove this a good time .. But no success. I don't know if is true. If it is, will help me a lot. Someone can give me a help with my doubt?

thanks in advance


No. Consider the open unit disk $\mathbb{D}$ in $\mathbb{R}^2$. Let $f_\infty$ be the characteristic function of some interval in $\partial\mathbb{D}$, say, of the upper hemisphere. Let $f_k:\partial\mathbb{D}\to\mathbb{R}$ be a monotone decreasing sequence of smooth functions converging to $f_\infty$ in an appropriate norm, say, a sequence of $\epsilon$-mollifications of $f_\infty$.

For each $k$, we have a harmonic function $u_k$ obtained by integrating the Poisson kernel $P$ against $f_k$. By the maximum principle, $0\leq u_{k+1}\leq u_k\leq 1$. These converge to $$u_\infty = \int_{\partial\mathbb{D}} f_\infty(\theta)P(\theta,z)\ d\theta$$ which extends to $f_\infty$ on $\partial\mathbb{D}$, which is discontinuous.

  • $\begingroup$ and if $f_k = f_{\infty}$ for all $k$, do you know something in this case? $\endgroup$ – math student Jun 11 '14 at 19:03
  • $\begingroup$ Then it's a constant sequence. $\endgroup$ – Neal Jun 11 '14 at 19:17
  • $\begingroup$ How can $f_k=f_{\infty}$ for all $k$? $f_\infty$ is discontinuous, whitle $f_k$ is smooth. $\endgroup$ – Tomás Jun 11 '14 at 19:53
  • $\begingroup$ you 're right guys . i dont know why i writed this . i did a great mystake haha . $\endgroup$ – math student Jun 11 '14 at 20:53

I don't think this is true. Consider, for example, the open cube $Q=(0,1)\times (0,1)$. Define on the boundary of $Q$ a function $v$ by the following: $v(x)=1$ for $x\in Q_1=\{(z,1):\ z\in [0,1]\}$ and $v(x)=0$ on the rest.

Consider the problem

$$\tag{1} \left\{ \begin{array}{rl} \Delta u=0 &\mbox{ if $x\in Q$} \\ u=v &\mbox{ if $x\in \partial Q$} \end{array} \right. $$

Problem $(1)$ has a unique solution $u$, which is harmonic in $Q$. Now, for each $n=1,2,...$, define a sequence of functions $v_n\in H^{1/2}(\partial Q)\cap C(\partial Q)$ by the following: $v_n(x)\le v_{n+1}(x)$,

$v_n(x)=1$ for $x\in \{(z,1):\ z\in [1/2^{n+1},1-1/2^{n+1}]\}$,

$0\le v(x)\le 1$ for $x\in \{(z,1):\ z\in [1/2^{n+2},1/2^{n+1}]\cup[1-1/2^{n+1},1-1/2^{n+2}]\}$,

$v_n(x)=0$ for $x\in \{(z,1):\ [0,1/2^{n+2}]\cup [1-1/2^{n+2},1]\}$.

For each $n$ there is a unique solution $u_n$ to the problem

$$ \left\{ \begin{array}{rl} \Delta u=0 &\mbox{ if $x\in Q$} \\ u=v_n &\mbox{ if $x\in \partial Q$} \end{array} \right. $$

The sequence $u_n$ is harmonic in $Q$ and is continous in $\partial Q$. By the maximum principle $u_n(x)\le u_{n+1}(x)$ for all $n$ and $u_n(x)\to u(x)$ for all $x\in \overline{Q}$.

Ps: It is possible that the above sequence $u_n$ is not continuous on the boundary, because $Q$ is not regular, however, this can be easily fixed, by considering regular sets, and applying the same reasoning as here, for example, one could consider regularize $Q$ on the corners, or one could consider a circle, and define $v$ to be different whether you take points in the upper half part or in the lower half part of the circle.

  • $\begingroup$ Your idea is good, but the OP wanted a decreasing, not increasing , sequence of functions. $\endgroup$ – user940 Jun 11 '14 at 0:54
  • $\begingroup$ Ops @ByronSchmuland, what a mistake. Thank you for correcting me. If I am not wrong, it can be corrected, by considering, $v=0$ in $Q_1$ and $1$ otherwise. Then, one can make the appropriate changes on $v_n$. $\endgroup$ – Tomás Jun 11 '14 at 10:52

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