Limit of Maximum of Continuous Complex Function

Question

See this MSE post for context. Let $G$ be a region/domain in $\Bbb{C}$ and $f : G \to \Bbb{C}$ continuous. Fix $z_0 \in G$. Why is it the case that

$$\lim_{z \to z_0} \max_{w \in [z,z_0]} |f(w)- f(z_0)|=0?$$

Intuitively, it makes sense. However, I am having trouble justifying it. Since $[z,z_0]$ is compact, for every $z$, there exists $t_z \in [0,1]$ such that $\max_{w \in [z,z_0]} |f(z)- f(z_0)| = |f((1-t_z)z + t_z z_0) - f(z_0)|$. As $z \to z_0$, why does $|f((1-t_z)z + t_z z_0) - f(z_0)| \to 0$? I tried proving that the assignment $z \mapsto t_z$ is continuous, but I didn't have much luck.

@copper.hat The limit is $0$ (I edited the post); and $w$ is a point on the line segment connecting $z$ and $z_0$. — user858792, Dec 14 '21 at 19:22
The variable $w$ does not appear in the formula so is pointless. — copper.hat, Dec 14 '21 at 19:24
That is a direct consequence of the continuity of $f$ at $z_0$. — Martin R, Dec 14 '21 at 19:37

score 1 · Answer 1 · answered Dec 14 '21 at 19:41

1

Try showing $\lim_{r \to 0} \sup_{w \in B(z_0,r)} |f(w)-f(z_0)| = 0$.

Hint:

Pick $\epsilon>0$. Since $f$ is continuous at $z_0$ there is some $\delta>0$ such that $|f(w)-f(z_0)| < \epsilon$ for all $w$ satisfying $|w-z_0| < \delta$.

answered Dec 14 '21 at 19:41

copper.hat

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Limit of Maximum of Continuous Complex Function

1 Answers1