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

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

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 $$\forall z_1,z_2\in\mathcal{Z}$$ $$$$|f(z_1,e_1)-f(z_2,e_2)|\le L\rho(z_1,z_2)+K\|e_1-e_2\|$$$$ where $$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)|.

• No you can't conclude continuity. You can't make $K \alpha$ small because it is fixed. Commented May 27, 2020 at 11:03
• I have included a modified version of the question. Can we now say that it is continuous and thus bounded due to EVT?
– amj
Commented May 28, 2020 at 5:19

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 $$L\mathrm{diam}\mathcal{Z} + K\alpha + |f(t_0)|$$.

• but what about $|f(t)|$. I mean it can still be unbounded?
– amj
Commented May 27, 2020 at 10:25
• If $t$ is fixed, this is a constant. Thus $f(z)$ is bounded by a constant not depending on $z$. Commented May 27, 2020 at 10:26
• I am sorry but I don't understand it. Is $t\in\mathcal{Z}$? then how can we say that $f(t)$ can't be unbounded? Please accept my apology if it should be self evident.
– amj
Commented May 27, 2020 at 10:30
• As $\mathcal{Z}$ is an abstract set, it does not have a particular point. So I choose $t \in \mathcal{Z}$. But if it was the ball of radius $1$, I would have choosen its center. I would then have shown that for any point $z$ in the ball, $|f(z)|$ would have been bounded by some constant, involving the diameter of the ball and the value of $f$ at its center. Here, $f(z)$ cannot be too far away from the value of $f$ at the particular point $t$ I choose before. Commented May 27, 2020 at 10:34
• I think I get it. I think. Thanks
– amj
Commented May 27, 2020 at 10:41