How to prove the inequality of the maximum function

Question

if we have a continuous real-valued function $f: I \to R$, where $I=([0,1] \times [0,1])$ and we have:

$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; |f(x,t)| \leq \epsilon$, for all $x,t\in I $,

then

$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; max|f(x,t)| \leq \epsilon$.

I think that this is true but how to prove such a thing. Any help would be appreciated.

Answer 1

If we assume that $\max\left\{\bigl\lvert f(x,y)\bigr\rvert\,\middle|\,(x,t)\in I^2\right\}$ exists, then the statement is trivial: this maximum is $\bigl\lvert f(x_0,t_0)\bigr\rvert$, for some $(x_0,t_0)\in I^2$. So,$$\max\left\{\bigl\lvert f(x,y)\bigr\rvert\,\middle|\,(x,t)\in I^2\right\}=\bigl\lvert f(x_0,t_0)\bigr\rvert\leqslant\varepsilon.$$

If you don't want to assume that $\max\left\{\bigl\lvert f(x,y)\bigr\rvert\,\middle|\,(x,t)\in I^2\right\}$ exists, then you can deduce it from the fact that $f$ is continuous and $I^2$ is compact.