TITLE: Show that $ h(x)=\max \{f(x,y) : y \in [0,1] \} $ is continuous
QUESTION [5 upvotes]: Let $ f(x,y) \ $ be a continuous real-valued function on the unit square $ [0,1] \times [0,1]$. 
Show that $$ h(x)=\max \{\,f(x,y) : y \in [0,1] \},  $$  is also continuous.
Answer.
Since   $ f(x,y)$ is continuous, then $ \max \{f(x,y) \}$ is also continuous on $[0,1 ] \times [0,1]$. 
Thus for any fixed values of $ y \in [0,1] \ $  ,  $ \max \{f(x,y) \}$  is also continuous .
i.e.,  $ \max \{f(x,y): y \in [0,1] \}=h(x) \ $ is also  continuous. 
But I need  a $ \varepsilon-\delta \ $ proof. Is there any?

REPLY [4 votes]: Since $f$ is continuous, then for every $\varepsilon>0$, there exists a $\delta>0$, such that
$$
|x_1-x_2|+|y_1-y_2|<\delta\quad\Longrightarrow\quad |\,f(x_1,y_1)-f(x_2,y_2)|<\varepsilon
$$ 
Consider $x_1,x_2\in[0,1]$, with $|x_1-x_2|<\delta$. Then there exists $y_1,y_2\in [0,1]$, such that
$$
|h(x_1)-h(x_2)|=
\Big|\max_y f(x_1,y)-\max_y f(x_2,y)\Big|=
| f(x_1,y_1)-f(x_2,y_2)|.
$$
Clearly, 
$0 \le f(x_1,y_1)-f(x_2,y_1)<\varepsilon\quad$ and
$\quad 0\le f(x_2,y_2)-f(x_1,y_2)<\varepsilon$.
Thus,
$$
f(x_1,y_1)<f(x_2,y_1)+\varepsilon\le f(x_2,y_2)+\varepsilon \tag{1}
$$
and
$$
f(x_2,y_2)<f(x_1,y_2)+\varepsilon\le f(x_1,y_1)+\varepsilon \tag{2}
$$
Combining (1) and (2), we obtain
$$
-\varepsilon< f(x_1,y_1)-f(x_2,y_2)<\varepsilon
$$
or equivalently
$$
|h(x_1)-h(x_2)|<\varepsilon.
$$