Show that g is continuous part 2

[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 posted this earlier with one proof, I am trying another more general

let ε >0 and arbitrary k. if f1(x) >...> fN(x) then since f1(x) - fN(x) is continuous, there is δ1 > 0 so that on |x-k| < δ1. we have that |f1(x) - fN(x)| > |f1(k) - fN(k)|/2

in particular then, for |x-k| <δ1 we have that max(f1(x),...,fN(x) = f(x). then there is δ2 > 0 so that for |x-k| < δ2 we have |f1(x)-f1(k)| <ε. now for |x-k| , δ = min(δ1, δ2) we have |max(f1(x),...,fN(x)) - max(f1(k),...,fN(k))| = |f1(x)-f1(k)| < ε.
The case of fN(k) >...>f1(k) is the same.

if f1(k) = fN(k) then there is δ1 >0 so that for |x-k| <δ1. |f1(x)-f1(k)| < ε and there is δ2 so that for |x-k| < δ2 we have |fN(x)-fN(k)|< ε. then for |x-k| < δ= min(δ1,δ2) we have
|max(f1(x),...,fN(x)) - max(f1(k),...,fN(k)| < ε.

since the above is either |f1(x)-f1(k)| or |fN(x)-fN(k)|.

by g(x) = max{f1(x),..., fN(x), then g(x) is also continuous

is this too general of a proof that is misses a lot in between?

 

[jbunniii]

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 posted this earlier with one proof, I am trying another more general

let ε >0 and arbitrary k. if f1(x) >...> fN(x) then since f1(x) - fN(x) is continuous, there is δ1 > 0 so that on |x-k| < δ1. we have that |f1(x) - fN(x)| > |f1(k) - fN(k)|/2

You need to be more careful about your strict inequalities. It may be that $f_1(x) = f_2(x) = \ldots = f_N(x)$, so your strict sequence of inequalities may not exist, even if you reorder the indexes.

Also, how does the $x$ in your $f_1(x) > f_2(x) > \ldots > f_N(x)$ relate to the $x$ in the rest of the sentence? I assume you recognize that for different values of $x$, the ordering may be different. For example, if $f_j(x) = jx$ for each $j \in \{1, \ldots, N\}$, then for $x > 0$ we have $f_N(x) > f_{N-1}(x) > \ldots > f_1(x)$, whereas for $x < 0$ the inequalities are reversed.

 

[ares25]

Im going to restructure for base step: some f(x) and g(x) that are continuous, then max(f(x),g(x)) is also continuous. use the proof above to prove that. switch out f1 for f(x) and fN for g(x). then say that from the base step f(x),g(x) we can have max f1,f2 is continuous then fi,fi+1 is also continuous up till N. so that we may may have max of each pairs up till N. since g(x) is the max of all pairs with each pair being continuous (by repetition of all max) g(x) is also continuous. Its rough but I hope I'm on the right process. Thanks again.

 

[jbunniii]

Im going to restructure for base step: some f(x) and g(x) that are continuous, then max(f(x),g(x)) is also continuous. use the proof above to prove that. switch out f1 for f(x) and fN for g(x). then say that from the base step f(x),g(x) we can have max f1,f2 is continuous then fi,fi+1 is also continuous up till N. so that we may may have max of each pairs up till N. since g(x) is the max of all pairs with each pair being continuous (by repetition of all max) g(x) is also continuous. Its rough but I hope I'm on the right process. Thanks again.

OK, if you post the details here after you finish the proof, I'll be happy to check it out.