# How prove this $g(x)=\sup{\{f(x,y)|0\le y\le 1\}}$ is continuous on $[0,1]$

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?

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

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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 $[0,1]$.

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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.

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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