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?

  • 1
    $\begingroup$ Do you happen to know about uniform continuity (on compact sets)? $\endgroup$
    – H. H. Rugh
    Oct 5, 2017 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 $y$? $\endgroup$ Oct 5, 2017 at 22:25

3 Answers 3


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 $$ |h(x_1)-h(x_2)|<\varepsilon. $$


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.


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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.