Section 3.2 The weak maximum principle
Theorem 3.2. Weak maximum principle.
Suppose that \(\Omega\) is bounded, \(L\) is uniformly elliptic in \(\Omega\text{,}\) and \(u \in C^2(\Omega) \cap C^0(\overline \Omega)\) satisfies \(Lu \ge 0\) in \(\Omega\text{.}\) If any of the conditions
- \(\displaystyle c \equiv 0\)
- \(c \le 0\) and \(\displaystyle\max_{\overline \Omega} u \ge 0\)
- \(\displaystyle \max_{\overline \Omega} u = 0\)
hold, then \(\displaystyle\max_{\overline \Omega} u = \max_{\partial \Omega} u
\text{.}\)
Proof.
First case: \(c \equiv 0\).
First we modify \(u\) slightly to make the inequality \(Lu \ge 0\) strict. For arbitrary \(\varepsilon \gt 0\) and a positive constant \(k\) to be determined later, define
\begin{equation*}
v(x) = u(x) + \varepsilon e^{kx_1}
\end{equation*}
Then
\begin{align*}
L[e^{kx_1}] \amp= \big(k^2a_{11}(x) + kb_1(x) \big) e^{kx_1}\\
\amp\ge \Big(\lambda_0k^2 - k\sup_{\Omega} |b_1| \Big)e^{kx_1}\\
\amp \gt 0,
\end{align*}
provided we choose \(k\) sufficiently large. Importantly, this choice does not depend on \(x\text{!}\) Hence we have \(Lv = Lu + \varepsilon Le^{kx_1} \gt 0\) in \(\Omega\text{.}\)
1
Since the dependence on \(k\) is so simple, we can even ‘solve’ the inequality to get an explicit threshold: \(k \gt \frac{1}{\lambda_0} \sup_{\Omega} |b_1|\text{.}\) In general, however, such algebraic manipulations are not always possible or worth the effort.
Next, assume for contradiction that \(\sup_\Omega v\) is attained at some point \(x_0 \in \Omega\text{.}\) Then by Proposition 2.34, \(\nabla v(x_0) = 0\) and the Hessian \(D^2v(x_0)\) is negative semi-definite, while the ellipticity of \(L\) implies that the matrix \(a(x_0)\) is positive semi-definite. Applying Lemma 2.31 with \(A=a\) and \(B=-D^2v(x_0)\text{,}\) we conclude that
\begin{align*}
Lv(x_0) \amp= a_{ij}(x_0)\partial_{ij}v(x_0) \le 0\text{.}
\end{align*}
But we showed earlier that \(Lv \gt 0\) in \(\Omega\text{,}\) so this is a contradiction.
Thus, for all \(\varepsilon \gt 0\text{,}\) \(v\) achieves \(\sup_\Omega v\) on the boundary \(\partial \Omega\text{.}\) For any \(x \in \overline \Omega\) we therefore have
\begin{align*}
u(x) \lt v(x) \amp\le \max_{\partial \Omega} v\\
\amp\le \max_{\partial \Omega} u + \varepsilon \max_{x \in \partial \Omega} e^{kx_1}.
\end{align*}
Sending \(\varepsilon \to 0\) we conclude that \(u(x) \le \max_{\partial \Omega} u\) as desired.
Second case: \(c \le 0\) and \(\max_{\overline\Omega} u \ge 0\).
We closely follow the proof of the previous case. By picking \(k \gt 0\) sufficiently large, we can still guarantee that \(Le^{kx_1} \gt 0\) in \(\Omega\text{.}\) Thus, for any \(\varepsilon \gt 0\text{,}\) the function \(v=u+\varepsilon e^{kx_1}\) satisfies \(Lv \gt 0\text{.}\) Moreover, \(\max_{\overline \Omega} v \gt
\max_{\overline \Omega} u\text{.}\) Suppose now that \(v\) achieves this maximum at an interior point \(x_0\text{.}\) As before we have \(a_{ij}(x_0)\partial_{ij}v(x_0) \le 0\text{.}\) Thus
\begin{gather*}
Lv(x_0) = a_{ij}(x_0)\partial_{ij}v(x_0) + c(x_0)\max_{\overline \Omega} v \le 0
\end{gather*}
where for the second term we have used the fact that \(c \le 0\) while \(\max_{\overline \Omega} v \gt 0\text{.}\) But this contradicts \(Lv \gt 0\text{.}\) The rest of the proof is identical.
Third case: \(\max_{\overline\Omega} u = 0\).
Decompose the zeroth-order coefficient \(c(x)\) as \(c = c^+ + c^-\) where \(c^+ = \max\{0,c\} \ge 0\) and \(c^- = \min\{0,c\} \le 0\text{.}\) Since \(\max_{\overline \Omega} u = 0\text{,}\) we have \(u \le 0\) and hence that \(c^+ u \le 0\text{.}\) Therefore
\begin{equation*}
0 \le Lu - c^+u = a_{ij} \partial_{ij}u + b_i \partial_i u+ c^-u = \tilde L u
\end{equation*}
where \(\tilde L = a_{ij} \partial_{ij} + b_i \partial_i + c^-\text{.}\) The result now follows by applying the previous case to the operator \(\tilde L\) and function \(u\text{.}\)
Remark 3.3. Minimum principles.
Replacing \(u\) by \(-u\) leads to equivalent ‘minimum principles’. More precisely, Theorem 3.2 continues to hold if
- \(Lu \ge 0\) is replaced by \(Lu \le 0\text{;}\)
- and \(\max\) is replaced by \(\min\) throughout; and
- the inequality \(\max_{\overline \Omega} u \ge 0\) in the second case is replaced by \(\min_{\overline \Omega} u \le 0\text{.}\)
Note carefully that the inequality \(c \le 0\) does not flip.
Proposition 3.4. Comparison principle for \(c \le 0\).
Let \(\Omega\) be bounded, \(L\) uniformly elliptic, \(c\le 0\text{,}\) and \(u,v \in C^2(\Omega)
\cap C^0(\overline \Omega)\text{.}\) If
\begin{align*}
\left\{
\begin{alignedat}{2}
L u \amp \le Lv \amp\quad\amp \ina \Omega, \\
u \amp \ge v \amp\quad\amp \ona \partial \Omega
\end{alignedat}
\right.
\end{align*}
then \(u\ge v\) in \(\Omega\text{.}\)
Proof.
Set \(w=v-u\text{.}\) If \(M = \max_{\overline\Omega} w \le 0\) we are done, so assume for the sake of contradiction that \(M \gt 0\text{.}\) By linearity \(Lw=Lv-Lu \ge 0\) in \(\Omega\text{,}\) and by assumption \(w \le 0\) on \(\partial\Omega\text{.}\) Applying the weak maximum principle, we have
\begin{gather*}
M = \max_{\overline \Omega} w = \max_{\partial \Omega} w \le 0\text{,}
\end{gather*}
which is a contradiction.
Corollary 3.5. Uniqueness of solutions for \(c\le 0\).
Let \(\Omega\) be bounded and let \(L\) be uniformly elliptic with \(c \le 0\text{.}\) If \(u,v \in C^2(\Omega) \cap C^0(\overline \Omega)\) satisfy \(Lu = Lv\) in \(\Omega\) and \(u=v\) on \(\partial \Omega\text{,}\) then \(u \equiv v\) in \(\Omega\text{.}\)
Proof.
As \(Lu \le Lv\) in \(\Omega\) and \(u \ge v\) on \(\partial\Omega\text{,}\) the comparison principle implies that \(u \ge v\) in \(\Omega\text{.}\) Similarly as \(Lu \ge Lv\) in \(\Omega\) and \(u \le v\) on \(\partial\Omega\text{,}\) the comparison principle implies that \(u \le v\) in \(\Omega\text{.}\) Thus \(u \equiv v\) as desired.
Definition 3.6. Sub- and supersolutions.
Consider a boundary value problem
\begin{equation}
\left\{
\begin{alignedat}{2}
L u \amp= f \amp\quad\amp \ina \Omega, \\
u \amp= g \amp\quad\amp \ona \partial \Omega.
\end{alignedat}
\right.\tag{3.3}
\end{equation}
We say that \(u \in C^2(\Omega) \cap C^0(\overline\Omega)\) is a supersolution of (3.3) if
\begin{equation*}
\left\{
\begin{alignedat}{2}
L u \amp \le f \amp\quad\amp \ina \Omega \\
u \amp \ge g \amp\quad\amp \ona \partial \Omega
\end{alignedat}
\right.
\end{equation*}
and a subsolution of (3.3) if
\begin{equation*}
\left\{
\begin{alignedat}{2}
L u \amp \ge f \amp\quad\amp \ina \Omega \\
u \amp \le g \amp\quad\amp \ona \partial \Omega
\end{alignedat}
\right.\text{.}
\end{equation*}
Note that being a solution of (3.3) is the same as being a supersolution and a subsolution simultaneously.
Exercises Exercises
1. Non-uniqueness.
(a)
Let \(\Omega = (0,\pi)^2 \subset \R^2\text{.}\) Show that \(u = \sin x_1
\sin x_2\) solves
\begin{align*}
\left\{
\begin{aligned}
\Delta u + 2u \amp = 0 \quad \text{ in } \Omega \\
u \amp = 0 \quad \text{ on } \partial\Omega
\end{aligned}
\right.
\end{align*}
conclude that this problem does not have a unique solution in \(C^2(\Omega)\cap
C^0(\overline\Omega)\text{.}\) Why does this not contradict Corollary 3.5?
Hint.
If you have any doubts about what the boundary of \(\Omega\) is, try drawing a picture.
Solution.
Certainly \(u \in C^2(\Omega)\cap C^0(\overline\Omega)\text{.}\) Moreover, on \(\partial\Omega\) we have either \(x_1 \in \{0,\pi\}\) or \(x_2 \in \{0,\pi\}\text{,}\) and hence \(u=0\text{.}\) To calculate \(\Delta u\) the summation convention is not particularly helpful; we find
\begin{align*}
\Delta u
\amp =
\partial_{11} (\sin x_1 \sin x_2)
+ \partial_{22} (\sin x_1 \sin x_2)\\
\amp =
\sin x_2 \partial_{11} \sin x_1
+ \sin x_1 \partial_{22} \sin x_2\\
\amp =
\sin x_2 \cdot (-\sin x_1)
+ \sin x_1 \cdot (-\sin x_2)\\
\amp = -2u\text{.}
\end{align*}
Thus the desired PDE holds.
To see that this problem does not have a unique solution, we must produce another one. The obvious choice is \(u \equiv 0\text{.}\) More generally, by linearity we have that \(\alpha u\) is a solution for any \(\alpha \in \R\text{.}\) Thinking back to linear algebra, we could call this a one-dimensional subspace of solutions.
Since the relevant elliptic operator \(L=\Delta + 2\) has zeroth-order coefficient \(c = 2\text{,}\) Corollary 3.5 does not apply; that result required \(c \le 0\) in \(\Omega\text{.}\)
Comment.
When calculating boundaries, it is often helpful to draw a picture. Some students in previous years have assumed there was a rule like
\begin{equation*}
\partial (A \times B) = (\partial A) \times (\partial B)\text{,}
\end{equation*}
which in this case told them that the boundary of the square \(\Omega\) was only its four corners. In this case the rule is instead
\begin{equation*}
\partial (A \times B) = (\partial A \times \overline B) \cup (\overline A \times \partial B)\text{,}
\end{equation*}
which will give you both the sides and the corners.
(b)
Give an example of an unbounded open set \(\Omega \subset \R^N\) for which the boundary value problem
\begin{align*}
\left\{
\begin{aligned}
\Delta u \amp = 0 \quad \text{ in } \Omega \\
u \amp = 0 \quad \text{ on } \partial\Omega
\end{aligned}
\right.
\end{align*}
does not have a unique solution \(u \in C^2(\Omega)\cap C^0(\overline\Omega)\text{.}\)
Hint.
There are many ways to do this. One option which has been popular in previous years is to look for monomials \(x^\alpha\) which satisfy \(\Delta x^\alpha = 0\text{,}\) and then to define \(\Omega\) by thinking about where these monomials vanish.
Solution.
Since \(u \equiv 0\) is always a solution, it suffices to find another solution \(u \not \equiv 0\text{.}\) One example is \(\Omega = (0,\infty) \times
\R^{N-1}\) and \(u = x_1\text{.}\) My personal favourite is to look at regions \(\Omega = (0,\pi) \times \R^{N-1}\) bounded by two parallel hyperplanes and then take a solution like \(u = \sin(x_1)e^{x_2}\) coming from separation of variables. Note that since the equation is linear, if \(u\) is a solution then so is \(\alpha u\) for any \(\alpha \in \R\) — as soon as we have two solutions we necessarily have uncountably many.
2. Subsolutions and supersolutions.
Let \(L\) be uniformly elliptic with \(c \le 0\) and \(\Omega\) be bounded, and fix \(f \in C^0(\overline\Omega)\) and \(g \in C^0(\partial\Omega)\text{.}\) Suppose that \(v \in C^2(\Omega) \cap C^0(\overline\Omega)\) is a supersolution of
\begin{align}
\left\{
\begin{alignedat}{2}
L u \amp= f \amp\quad\amp \ina \Omega, \\
u \amp= g \amp\quad\amp \ona \partial \Omega.
\end{alignedat}
\right.\tag{✶}
\end{align}
and that \(w \in C^2(\Omega) \cap C^0(\overline\Omega)\) is a subsolution of (✶).
(a)
Show that \(w \le v\text{.}\)
(b)
If \(u \in C^2(\Omega) \cap C^0(\overline\Omega)\) is solution of (✶), show that \(w \le u \le v\text{.}\)
3. (PS4) A two-sided estimate.
Suppose that \(B_r(p) \subseteq \Omega \subseteq B_R(q)\text{,}\) and that \(f\maps \Omega \to \R\) satisfies
\begin{equation*}
0 \le m \le f(x) \le M
\end{equation*}
for all \(x \in \Omega\text{.}\)
(a)
Prove that, if it exists, the solution \(u \in C^2(\Omega)
\cap C^0(\overline{\Omega})\) of
\begin{align*}
\left\{
\begin{alignedat}{2}
-\Delta u \amp= f \amp\quad\amp \ina \Omega, \\
u \amp= 0 \amp\quad\amp \ona \partial \Omega
\end{alignedat}
\right.
\end{align*}
satisfies the two-sided estimate
\begin{equation*}
\frac m {2N} (r^2 -\abs{x-p}^2)
\le
u(x)
\le
\frac M{2N} (R^2 -\abs{x-q}^2).
\end{equation*}
Hint.
Use the comparison principle. Note that while the operator \(\Delta\) is elliptic, the operator \(-\Delta\) is not.
Solution.
We will use the comparison principle with the comparison functions
\begin{align*}
u_1 \amp= \frac m {2N} (r^2 -\abs{x-p}^2),\\
u_2 \amp= \frac M{2N} (R^2 -\abs{x-q}^2),
\end{align*}
which are of course defined and smooth on all of \(\R^N\) and hence in particular on \(\Omega\text{.}\) We calculate \(\Delta u_1 = -m\) and \(\Delta u_2 = -M\) (again on all of \(\R^N\) and hence on \(\Omega)\text{,}\) and so
\begin{gather*}
\Delta u_2 \le \Delta u \le \Delta u_1 \ina \Omega.
\end{gather*}
In order to conclude the desired inequalities
\begin{gather}
u_1 \le u \le u_2 \ina \Omega\tag{✶}
\end{gather}
from the comparison principle, we need to show
\begin{gather}
u_1 \le u \le u_2 \ona \partial\Omega.\tag{✶✶}
\end{gather}
Since \(B_r(p) \subseteq \Omega\) we have \(\abs{x-p} \ge r\) on \(\partial\Omega\text{,}\) and hence \(u_1
\le 0 = u\text{.}\) Similarly \(\Omega \subseteq B_R(q)\) implies \(\abs{x-q} \le R\) on \(\partial\Omega\) and hence \(u_2 \ge 0 = u\text{.}\) Thus (✶✶) holds and we can indeed deduce (✶) as desired.
Comment 1.
In the notation of the official solution, many students correctly observed that \(u_1 = 0\) on \(\partial B_r(p)\) and \(u_2 = 0\) on \(\partial
B_R(q)\text{.}\) Note, though, that these equalities are not what the comparison principle is asking for. It wants inequalities \(u_1 \le u \le
u_2\) on \(\partial\Omega\text{,}\) which are a bit more subtle.
It may be clarifying to sketch a cartoon of the set \(\Omega\) and the inclusions \(B_r(p) \subseteq \Omega \subseteq B_R(q)\)
Comment 2.
While \(\Delta\) is elliptic, \(-\Delta\) is not. If you set \(L=-\Delta\text{,}\) you will have to be quite careful with your inequalities when referencing results from lectures (some of them will flip).
(b)
Can you choose \(u,f,\Omega,r,R\) so that either of these inequalities is an equality?
Solution.
Let \(u_1,u_2\) be as in the previous part. A cheeky way to respond is to say yes, consider the case where \(m=M=0\) so that \(f,u,u_1,u_2\equiv
0\text{.}\) A slightly less cheeky example would be to take \(m=M \ne 0\) (so \(f\) is constant) and \(p=q\) and \(r=R\) so that \(\Omega\) is a ball. Then we can easily check that \(u_1 = u = u_2\) in all of \(\Omega\text{.}\)
4. (PS5) Domains contained in a slab.
Let \(\Omega\) be bounded and contained in the slab \(0\le x_1\le d\text{,}\) let \(L\) be uniformly elliptic with \(c \le 0\text{,}\) and let \(u \in C^2(\Omega) \cap
C^0(\overline\Omega)\text{.}\) Finally, let \(f=Lu\text{,}\) and suppose that \(f\) is a bounded function.
(a)
Show that there is a constant \(k \gt 0\) depending only on the ellipticity constant \(\lambda_0\) of \(L\) and on \(\sup_\Omega \abs{b_1}\) so that the auxiliary function \(z(x)=e^{kd}-e^{kx_1}\) satisfies \(Lz \le -1\) in \(\Omega\text{.}\)
Hint.
This is similar to an argument made in the proof of Theorem 3.2. Since \(c \le 0\) and \(z \ge 0\) in \(\Omega\text{,}\) the \(cz\) term is harmless.
Solution.
\begin{align*}
L z
\amp = -k (k a_{11} + b_1) e^{kx_1} + cz\\
\amp \le -k (k a_{11} + b_1) e^{kx_1}
\end{align*}
in \(\Omega\text{.}\) By uniform ellipticity the first two factors above satisfy
\begin{gather*}
k (k a_{11} + b_1)
\ge k \Big(k\lambda_0 - \sup_\Omega \abs{b_1}\Big)
\ge 1
\end{gather*}
in \(\Omega\text{,}\) provided we choose \(k \gt 0\) sufficiently large depending on \(\lambda_0\) and \(\sup_\Omega\abs{b_1}\text{,}\) for instance
\begin{equation*}
k = \max\Big(1, \frac 1{\lambda_0} \sup_\Omega\abs{b_1}\Big)\text{.}
\end{equation*}
Since \(x_1 \ge 0\) in \(\Omega\text{,}\) inserting this into our earlier estimate for \(Lz\) yields
\begin{align*}
L z
\amp \le -e^{kx_1}
\le -1
\end{align*}
in \(\Omega\) as desired.
Comment 1.
The above solution gives an explicit formula for \(k\text{,}\) but I want to emphasise that this is not strictly necessary. What we are using here is the following elementary fact: If
\begin{equation*}
p(k)=p_0+p_1 k + \cdots + p_n k^n
\end{equation*}
is a polynomial with coefficients \(p_0,\ldots,p_{n-1} \in \R\) and \(p_n \gt
0\text{,}\) then \(\lim_{k \to \infty} p(k) = \infty\text{.}\) Here our polynomial is
\begin{equation*}
p(k) = k^2 \lambda_0 - \sup_\Omega \abs{b_1} k
\end{equation*}
with \(n=2\) and \(p_2 = \lambda_0 \gt 0\text{.}\) Without even thinking about the quadratic formula, we know that \(p(k) \to \infty\) as \(k \to \infty\text{.}\) In particular, there exists \(K \gt 0\) depending only on \(\lambda_0,\sup_\Omega
\abs{b_1}\) such that \(k \ge K\) implies \(p(k) \ge 1\text{.}\)
Comment 2.
Note that we need \(Lz \le -1\) here, and not just \(Lz \le 0\text{.}\) There are many ways to accomplish this, but in the official solutions the strategy is to
- use the inequality \(e^{kx_1} \ge 1\text{,}\) which holds in \(\Omega\) as \(x_1,k \ge 0\text{;}\) and
- arrange for the remaining factor (up to a sign) to satisfy \(k (k a_{11} + b_1) \ge 1\) in \(\Omega\text{.}\)
If either of these \(\ge 1\) bounds is replaced by \(\ge 0\text{,}\) then the argument fails.
(b)
With \(z\) as above, set
\begin{equation*}
v = \Big(\sup_\Omega \abs{f}\Big) z + \max_{\partial\Omega} \abs u\text{.}
\end{equation*}
By applying the comparison principle to \(u\) and \(\pm v\text{,}\) show that \(\abs u \le v\) in \(\Omega\text{.}\)
Hint.
Note that \(-\sup_\Omega \abs f \le f \le \sup_\Omega \abs f\) in \(\Omega\text{,}\) and similarly for \(u\) on \(\partial\Omega\text{.}\)
Solution.
Since \(c \le 0\) and \(\max_{\partial\Omega} \abs u \ge 0\text{,}\) the previous part gives
\begin{align*}
L v
\amp = \Big(\sup_\Omega f\Big) Lz + c \max_{\partial\Omega} \abs u\\
\amp \le -\sup_\Omega \abs f \\
\amp \le f = Lu
\end{align*}
in \(\Omega\text{.}\) On \(\partial\Omega\) we have \(z \ge 0\) and hence \(v \ge \max_{\partial\Omega} \abs u \ge u\text{.}\) Applying the comparison principle, we deduce that \(v \ge u\) in \(\Omega\text{.}\) Arguing similarly with \(v\) replaced by \(-v\) we get \(-v \le u\) in \(\Omega\text{,}\) and hence \(\abs
u \le v\) as desired.
Comment.
Be careful not to immediately discard constant terms when applying differential operators. Here, for instance, we have
\begin{align*}
Lv \amp= L\left( \Big(\sup_\Omega f\Big) z + \max_{\partial\Omega} \abs u \right)\\
\amp= \Big(\sup_\Omega f\Big) Lz + \Big(\max_{\partial\Omega} \abs u\Big) L1\\
\amp= \Big(\sup_\Omega f\Big) Lz + \Big(\max_{\partial\Omega} \abs u\Big) c\text{.}
\end{align*}
(c)
Conclude that there exists a constant \(C\) depending only on \(\lambda_0\text{,}\) \(d\) and \(\sup_\Omega \abs{b_1}\) so that
\begin{equation*}
\max_{\overline \Omega} \abs u \le \max_{\partial\Omega} \abs u + C \sup_\Omega \abs{f}.
\end{equation*}
Solution.
By the previous part we have
\begin{align*}
\max_{\overline\Omega} \abs u
\amp \le \max_{\overline\Omega} v \\
\amp = \max_{\partial\Omega} \abs u + \Big(\max_{\overline \Omega} z\Big)\Big(\sup_\Omega \abs f\Big)\\
\amp \le \max_{\partial\Omega} \abs u + (e^{kd}-1)\sup_\Omega \abs f,
\end{align*}
which is an inequality of the desired form with \(C = e^{kd}-1\text{.}\) Finally, we observe that \(C\) depends only on \(d\) and on \(k\text{,}\) and that \(k\) in turn depends only on \(\lambda_0\) and \(\sup_\Omega \abs{b_1}\text{.}\)
Comment.
Note that we cannot quite set \(C = \max_{\overline\Omega} z\text{,}\) as that would depend not only on \(\lambda_0,d,\sup_\Omega \abs{b_1}\) but also on the domain \(\Omega\) itself.
(d)
Suppose that we have a sequences \(u_n \in C^2(\Omega) \cap C^0(\overline\Omega)\) and \(f_n \in C^0(\overline\Omega)\) with \(Lu_n = f_n\text{,}\) and such that \(f_n \to f\) in \(C^0(\overline\Omega)\) and \(u_n \to u\) in \(C^0(\partial\Omega)\text{.}\) By applying the previous part to \(u_n-u\text{,}\) show that \(u_n \to u\) in \(C^0(\overline\Omega)\text{.}\)
Solution.
Applying the previous part to \(u_n-u\text{,}\) which has \(L(u_n-u)=f_n-f\text{,}\) we have
\begin{align*}
\n{u_n-u}_{C^0(\overline\Omega)}
\amp =
\max_{\overline\Omega} \abs{u_n-u}\\
\amp \le
\max_{\partial\Omega} \abs{u_n-u} + C \sup_\Omega \abs{f_n-f}\\
\amp =
\n{u_n-u}_{C^0(\partial\Omega)} + C \n{f_n-f}_{C^0(\overline\Omega)}\\
\amp \to 0
\end{align*}
as \(n \to \infty\text{.}\)
Comment.
Let \(\Omega\) be bounded and \(f_n,f \in C^0(\overline \Omega)\text{.}\) In part thanks to Exercise 2.8.1, the following statements are equivalent
- \(f_n \to f\) uniformly on \(\Omega\)
- \(f_n \to f\) uniformly on \(\overline \Omega\)
- \(\displaystyle \sup_\Omega \abs{f_n - f} \to 0\)
- \(\displaystyle \sup_{\overline \Omega} \abs{f_n - f} \to 0\)
- \(\displaystyle \max_{\overline \Omega} \abs{f_n - f} \to 0\)
You are free to use this fact without comment, and certainly without looking up the number of Exercise 2.8.1.
