Show that a function is continuous in order to use extreme value theorem

Question

Let $f:\mathcal{Z}\times\mathcal{Z}\rightarrow \mathbb{R}$ be a function defined on a compact metric space $(\mathcal{Z},\rho)$ such that $\forall z_1,z_2\in\mathcal{Z}$ \begin{equation} \label{eq1} |f(z_1)-f(z_2)|\le L\rho(z_1,z_2)+K\alpha \end{equation} where $L,K$ and $\alpha$ are constants.

I want to show that $f$ is bounded from above. Now, Extreme Value theorem can be used to show that a continuous function on a compact space is bounded.

Can I say that $f$ is continuous because for all $z_1,z_2\in\mathcal{Z}, \epsilon>0$, whenever $\rho(z_1,z_2)<\epsilon$, $|f(z_1)-f(z_2)|<L\epsilon+K\alpha$.

or Let's look at a modified version of this question:

Let $f:\mathcal{Z}\times\mathcal{Z}\rightarrow \mathbb{R}$ be a function defined on a compact metric space $(\mathcal{Z},\rho)$ such that $\forall z_1,z_2\in\mathcal{Z}$ \begin{equation} |f(z_1,e_1)-f(z_2,e_2)|\le L\rho(z_1,z_2)+K\|e_1-e_2\| \end{equation} where $L,K$ and $\|e_1-e_2\|\leq\alpha$ are constants.

I want to show that $f$ is bounded from above. Now, Extreme Value theorem can be used to show that a continuous function on a compact space is bounded.

Can I now say that $f$ is continuous because for all $z_1,z_2\in\mathcal{Z}, \epsilon>0$, whenever $\rho(z_1,z_2)<\epsilon$, $|f(z_1)-f(z_2)|<L\epsilon+K\alpha$.

Answer 1

It can be discontinuous. For example, chose a particular point $x$, and define $f(z)$ to be $K\alpha$ if $z = x$, $0$ if not. Then this is discontinuous but for all $z_1$, $z_2$, you have $|f(z_1)-f(z_2)| \leqslant K\alpha \leqslant L\rho(z_1,z_2) + K\alpha$.

What is true is that $f$ is bounded because for $z,t_0 \in \mathcal{Z}$ \begin{align} |f(z)| &=|f(z)-f(t_0) + f(t_0)| \\ & \leqslant |f(z)-f(t_0)| + |f(t_0)| \\&\leqslant L \rho(z,t_0) + K\alpha + |f(t_0)| \\&\leqslant L\mathrm{diam}\mathcal{Z} + K\alpha + |f(t_0)| \end{align} (recall that $\mathcal{Z}$ is compact thus as a finite diameter.) So if $t$ is fixed, $f$ is bounded by the constant $L\mathrm{diam}\mathcal{Z} + K\alpha + |f(t_0)|$.