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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?

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    $\begingroup$ Do you happen to know about uniform continuity (on compact sets)? $\endgroup$ – H. H. Rugh Oct 5 '17 at 8:25
  • $\begingroup$ max${f(x,y)}$ would need some clarification on meaning because I dont think your 'answer' correct. Is it a constant function in $x$ and $y$? $\endgroup$ – Calvin Khor Oct 5 '17 at 22:25
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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 $$ |h(x_1)-h(x_2)|<\varepsilon. $$

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You need to use this inequality.

$$|max \ f(x) - max \ g(x)| \le max|f(x) - g(x)|$$

Then, write the difference below.

$$|h(x)-h(x_0)|=|max\{f(x,y):y\in [0,1]\}-max\{f(x_0,y):y\in [0,1]\}|$$

Note that the two maximum functions, in the expression above, can be considered as $f(x)$ and $g(x)$ that we used in the inequality. So, using the inequality, the expression becomes

$$|max\{f(x,y):y\in [0,1]\}-max\{f(x_0,y):y\in [0,1]\}| \leq max\{|f(x,y)-f(x_0,y)|:y\in [0,1]\}$$

then, we use the continuity of $f(x,y)$ to say

$$\forall\epsilon \hspace{1mm}\exists \hspace{1mm} \delta|\hspace{1mm} |d(x-x_0,y-y)|=|d(x-x_0,0)|=|x-x_0|<\delta \implies |f(x,y)-f(x_0,y)|<\epsilon$$

Now, you need to choose the delta to be as below, knowing that for a data point $(x,y)$, $\delta$ is a function of $\epsilon$. So write $\delta(\epsilon,x,y)$

$$\delta(\epsilon) = inf\{\delta(\epsilon,x,y)|x,y\}$$

with this delta, you can go in the inverse direction.

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Suppose that $h$ is not continuous at some point $x_0.$ then there exists a number $\varepsilon >0$ and a sequence $x_n \to 0 $ such that $$|h(x_n ) -h(x_0 )|\geq \varepsilon.$$ Since the functions $\xi_n (y) =f(x_n ,y ) $ are continuous on $[0,1],$ thus for any $n$ there exists $y_n $ such that $h(x_n )= \xi_n (y_n) =f(x_n , y_n ).$ But the interval $[0,1] $ is compact therefore $y_{n_j} \to y_0 \in [0,1],$ for some sequence $(n_j ).$ Hence $$f(x_{n_k} , y^{\chi} )-f(x_0 ,y^{\chi})\leq f(x_{n_k} , y_{n_k} )-f(x_0 ,y^{\chi})\leq -\varepsilon \vee f(x_{n_k} , y^{\chi} )-f(x_0 ,y^{\chi})\geq f(x_{n_k} , y_{n_k} )-f(x_0 ,y^{\chi})\geq \varepsilon$$ where $y^{\chi}$ is such that $h(x_0 ) = f(x_0 ,y^{\chi} ) .$ But the last implies that $$0=f(x_0 , y^{\chi} )-f(x_0 ,y^{\chi})\leq f(x_0 , y_0 )-f(x_0 ,y^{\chi})\leq -\varepsilon \vee 0=f(x_0 , y^{\chi} )-f(x_0 ,y^{\chi})\geq f(x_0 , y_0 )-f(x_0 ,y^{\chi})\geq \varepsilon$$ Contradiction

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