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Let $f$ be continuous on a domain (open and connected), $D$. Now suppose that for every $z_0 \in D$, there exists $B_\delta (z_0)$ such that $f(z)$ is constant in $B_\delta (z_0)$. Show that $f$ is constant in $D$.

I don't know if this is correct, but allow me to share my thoughts. If $f$ is continuous on $D$, then $f$ is continuous on every $z_0 \in D$. Define an open ball, $B_\delta (z_0)$, such that $f(z) = K$, for all $z \in B_\delta (z_0)$.

Since $D$ is a region, it is connected, then $D\ne S_1 \cup S_2$, where $S_1 \cap S_2 = \emptyset$. Then, $D$ is the union of open balls that are not disjoint. Hence $f(z)$ is constant in $D$.

Martin Sleziak
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Harry Evans
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2 Answers2

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Let $z_0 \in O$ and define $c = f(z_0)$.

Then define $U = \{x \in O: f(x) = c\}$.

Then:

  1. $U \neq \emptyset$, as witnessed by $z_0$.

  2. $U$ is open: if $z \in U$, then we have $\delta > 0$ such that $f$ is constant on $B(z,\delta)$. As $z \in U$ already tells us that $f(z) = c$ all values of $f$ on this ball $B(z,\delta)$ are $c$ too. So $B(z,\delta) \subset U$ and so $z$ is an interior point of $U$.

  3. $U$ is closed in $O$: suppose we have $z \in O$ such that $z \in \overline{U}$. Let $\delta>0$ again be such that $f$ restricted to $B(z,\delta)$ is constant. Then as $z \in \overline{U}$, there is some $p \in U$ with $p \in B(z,\delta)$. So $f(p) = c$ and as $f$ has all the same values for any point of $B(z,\delta)$, we know that $f(z) = c$ as well. So by definition, $z \in U$, so $\overline{U} \subseteq U$ so $U$ is closed (every limit point of $U$ is already in $U$).

As $O$ is connected we knwo that any closed and open set that is non-empty must equal $O$. So $O = U$, and $f$ only assumes the value $c$. Done.

Henno Brandsma
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By the assumptions, $D$ must be path connected. So let $x,y \in D$ and $\gamma : [0,1] \to D$ be a path connecting them. Then the image of $\gamma$ is compact and hence there are $x_1,\cdots,x_n \in D$ such that $$ \mathrm{Im}(\gamma) \subset \bigcup_{i=1}^n B(x_i,\delta_i), $$ where $f$ is constant on each of the $B(x_i,\delta_i)$. Now you see that $f(x) = f(y)$.

  • how come the continuity of f is not used in the proof? Does this mean that continuity is not required? – Harry Evans Jan 28 '17 at 10:43
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    @HarryEvans continuity is sort of superfluous. A locally constant function is automatically continuous, and locally constant is what is used in the proof (so the stronger notion). – Henno Brandsma Jan 28 '17 at 13:08