Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

let $f(x,y):[0,1]\times[0,1]\to R$ is continuous real function.

show that $$g(x)=\sup{\{f(x,y)|0\le y\le 1\}}$$ is continuous on $[0,1]$

My try: since $f(x,y)$ is continuous on $D=[0,1]\times [0,1]$, so $f(x,y)$ is Uniformly continuous on $D$,so $\forall\varepsilon>0$,then exist $\delta>0$,such $|x_{1}-x_{2}|<\delta,|y_{1}-y_{2}<\delta$,then we have $$|f(x_{1},y_{1})-f(x_{2},y_{2})|<\varepsilon$$ so $$g(x_{1})-g(x_{2})=|\sup f(x_{1},y)-\sup f(x_{2},y)|<\sup|f(x_{1},y)-f(x_{2},y)|$$

Now maybe follow is not true?

$$|\sup f(x_{1},y)-\sup f(x_{2},y)|<\sup|f(x_{1},y)-f(x_{2},y)|$$

share|cite|improve this question
up vote 1 down vote accepted

Since $f(x,y)$ is continuous on $D=[0,1]\times [0,1]$, so $f(x,y)$ is uniformly continuous on $D$,so for a given $\varepsilon>0$,there exsits $\delta>0$, s.t. $$|f(x_{1},y_{1})-f(x_{2},y_{2})|<\frac{\varepsilon}{2}$$whenever $\sqrt{(x_{1}-x_{2})^2+(y_{1}-y_{2})^2}<\delta$.

Given $x_0\in[0,1]$, for all $x$ satisfy $|x-x_0|<\delta$ and $x\in[0,1]$, we have\[f(x,y)\leqslant f(x_0,y)+\frac{\varepsilon}{2}\leqslant g(x_0)+\frac{\varepsilon}{2}\Longrightarrow g(x) < g(x_0)+\varepsilon.\]Similarly, we have\[g(x_0) < g(x)+\varepsilon.\]Consequently, $|g(x_0)-g(x)|<\varepsilon$ whenever $|x-x_0|<\delta$, which implies $g(x)$ is continuous at $x=x_0$. As $x_0$ is choosed randomly, so $g(x)$ is continuous on $[0,1]$.

share|cite|improve this answer

Let $x \in [0,1]$ be given. Then $\sup_{0 \le y \le 1}f(x,y)=f(x,y_{x})$ for some $y_{x}$ because, for a fixed x, the function $y\mapsto f(x,y)$ is continuous and, hence, achieves its maximum value.

Let $\epsilon > 0$ be given. Because $f$ is uniformly continuous on $[0,1]\times[0,1]$, there exists $\delta > 0$ such that $|f(x,y)-f(x',y')| < \epsilon/2$ for any points $(x,y), (x',y') \in [0,1]\times[0,1]$ for which $|x-x'| < \delta$ and $|y-y'| < \delta$. It follows that, if $|x-x'| < \delta$, one has $|f(x,y_{x})-f(x',y_{x})| < \epsilon$, thereby guaranteeing that $$ \sup_{0\le y\le 1}f(x',y) \ge f(x',y_{x}) > f(x,y_{x})-\epsilon = \sup_{0\le y\le 1}f(x,y)-\epsilon. $$ Likewise, for $|x-x'| < \delta$, $$ \sup_{0 \le y \le 1}f(x,y) > \sup_{0\le y\le 1}f(x',y)-\epsilon. $$ So, whenever $|x-x'| < \delta$, $$ \sup_{0\le y \le 1}f(x',y)+\epsilon > \sup_{0\le y\le 1}f(x,y) > \sup_{0\le y \le 1}f(x',y)-\epsilon, $$ $$ |\sup_{0\le y \le 1}f(x,y)-\sup_{0\le y\le 1}f(x',y)| < \epsilon. $$ Because $\epsilon > 0$ was arbitrary, then it follows that $x\mapsto \sup_{0\le y\le 1}f(x,y)$ is a uniformly continuous function of $x$ on $[0,1]$ and, hence, is continuous.

share|cite|improve this answer
How does the "Likewise" work exactly? – ronno Dec 16 '13 at 13:31
swap x and x'. There's nothing distinguishing the two. – TrialAndError Dec 16 '13 at 17:30
Of course there is, the $\delta$ depends on the $y_x$ which depends on $x$. I believe the statement crucially depends on the uniform continuity of $f$, but I don't have a counterexample. – ronno Dec 16 '13 at 17:31
Swap x and x' in the argument from the beginning. They start off indistinguishable from each other because of uniform continuity. When you do swap the two, you'll use $y_{x'}$ instead of $y_{x}$, and the argument gives the first inequality where x' and x are interchanged. – TrialAndError Dec 16 '13 at 17:38
But the $\delta$ need not be the same. $x$ need not be within $\delta'$ of $x'$. So $g(x) \ge g(x')-\epsilon/2$ need not hold. – ronno Dec 16 '13 at 17:39

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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