# Show that g is a continious function

1. May 12, 2014

### ares25

Let f1,....,fN be continuous functions on interval [a,b]. Let g:[a,b] -> R be the function give by
g(x) = max{ f1(x),...., fN(x)}.

show that g is a continuous function.

I have let t ε [a,b]
NTS g(x) is continous at t
if a≤t< b, then there exists a neighborhood Vt of t contained in [a,t), such that g(x) = f1(x) for any x ε Vt.
Note that f1(x) is continuous on [a,b]. we obtain f1 is continuous on t. → g(x) is is continuous at t.
likewise, if a<t≤b, then there exists a neighborhood Vt of t contained in (t,b], such that g(x) = fN(x) for any x ε Vt.
Note that fN is continuous on [a,b]. fN is continuous on t. g(x) is continuous at t.

by g(x) = f1(x) True, then g(x) = fi+1(x) is also true for all i ε N.

IS this getting close or in the right ballpark? because i also have another more general one.

2. May 12, 2014

### jbunniii

How do you figure that? What is special about $f_1$ that would ensure that it is the smallest of the $f_n$'s on this interval, or on any interval?

Here is a hint which may help. Note that for $N > 1$, we have $\max\{f_1(x), f_2(x), \ldots, f_N(x)\} = \max\{\max\{f_1(x), f_2(x), \ldots f_{N-1}(x)\}, f_N(x)\}$. So this suggests that an inductive proof could work.

Another hint: $\max\{a, b\} = (a + b + |a-b|) / 2$.

3. May 12, 2014

### ares25

On my other proof https://www.physicsforums.com/showthread.php?t=753405 I made it more general for the whole, but i believe its too general. I used that idea of max for whole but want to switch it to inductive. where the max of f,g is also continuous, then the max of any two are also continuous, and therefore we can show it for any fn+1. if you could take a look at that proof, it be great help. I'm still breaking it down but thanks for the advice.