3

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?

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

math110
  • 93,304

2 Answers2

1

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 $[0,1]$.

Nirvanacs
  • 1,523
  • 9
  • 14
1

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.

Disintegrating By Parts
  • 87,459
  • 5
  • 65
  • 149
  • swap x and x'. There's nothing distinguishing the two. – Disintegrating By Parts Dec 16 '13 at 17:30
  • 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. – Disintegrating By Parts Dec 16 '13 at 17:38
  • No, the delta is the same. You get |f(x,y)-f(x',y')| < epsilon whenever |x-x'| < delta and |y-y'| < delta. Nothing preferential in that statement. The delta is not chosen according to x,y or according to x',y'. The delta in independent of the choice of (x,y) and (x',y'), so long as |x-x'| and |y-y'| are small. – Disintegrating By Parts Dec 16 '13 at 17:43