Continuity Proof

May 2010
11
0
Sorry reposted:

Let f(x) and g(x) be fcns de fned on an interval I and let h(x) = max(f(x); g(x)) and j(x) = min(f(x); g(x)). h(x) and
j(x) are continuous on I. Show f(x) and g(x) are continuous on I or give a counterexample.
 
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Drexel28

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Nov 2009
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Berkeley, California
Let f(x) and g(x) be functions defi ned on an open interval I and let h(x) = max{f(x); g(x)} and j(x) = min{f(x); g(x)}. Suppose that h(x) and
j(x) are continuous on I. Either show that f(x) and g(x) are also continuous on I or else give a counterexample.


Any ideas on how to structure this proof using contradiction?

Thanks!
How much do you know about analysis?
Hint:

\(\displaystyle \left(f\vee g\right)^{-1}((a,\infty))=\left\{x\in I:f(x)>a\right\}\cup\left\{x\in I:g(x)>a\right\}\)
 
May 2010
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I know analysis at an introduction course level. I know about delta-epsilon, continuity definitions, IVT, etc.
 

Drexel28

MHF Hall of Honor
Nov 2009
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I know analysis at an introduction course level. I know about delta-epsilon, continuity definitions, IVT, etc.
Hmm. Alright, so I guess the easiest way to say it then would be that \(\displaystyle \max\{f,g\}=\frac{f+g+|f-g|}{2}\), can you show that's continuous by using some of your known theorems?
 
May 2010
11
0
Well, we know that the function h(x) is continuous, so the max(f,g) would be continuous. I know that there are rules for algebraic combinations of continuous functions being continuous, so I'm wondering if the max function can be worked backwards to prove f and g continuous that way. Is that where you were going with this?
 

Drexel28

MHF Hall of Honor
Nov 2009
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Berkeley, California
Well, we know that the function h(x) is continuous, so the max(f,g) would be continuous. I know that there are rules for algebraic combinations of continuous functions being continuous, so I'm wondering if the max function can be worked backwards to prove f and g continuous that way. Is that where you were going with this?
Oh, wow! I'm sorry, I completely misread the question. Please forgive me, I have to go. I hope another member can help you. Once again, I'm sorry (Worried)
 

Opalg

MHF Hall of Honor
Aug 2007
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Leeds, UK
Let f(x) and g(x) be functions defi ned on an open interval I and let h(x) = max{f(x); g(x)} and j(x) = min{f(x); g(x)}. Suppose that h(x) and
j(x) are continuous on I. Either show that f(x) and g(x) are also continuous on I or else give a counterexample.


Any ideas on how to structure this proof using contradiction?
A better idea would be to look for a counterexample, perhaps by taking f to be a function that only takes the values 0 and 1, with a jump discontinuity at some point. Then think about how to construct g so as to make h and j continuous.
 
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Apr 2010
3
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Let f(x) and g(x) be functions defi ned on an open interval I and let h(x) = max{f(x); g(x)} and j(x) = min{f(x); g(x)}. Suppose that h(x) and
j(x) are continuous on I. Either show that f(x) and g(x) are also continuous on I or else give a counterexample.


Any ideas on how to structure this proof using contradiction?

Thanks!
take two parts, first suposse than \(\displaystyle x\) (element of domain) is point internal, and demostrate than is continiuos

second suposse that \(\displaystyle x\) satisfy f(x)=g(x) and probe that is continius.
 
Jan 2010
150
29
Mexico City
Or simply take these two easy functions on [0,1]


f(x)=1 if \(\displaystyle 0\leq x<\frac{1}{2} \)
-1 if \(\displaystyle \frac{1}{2} \leq x \leq 1\)


g(x)= -1 if \(\displaystyle 0\leq x <\frac{1}{2}\)
1 if \(\displaystyle \frac{1}{2} \leq x \leq1\)


both are discontinuous but max and min functions h,j are constant hence continuous
 
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Mar 2009
419
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Uptown Manhattan, NY, USA
Uh, don't delete the question just because it has been answered.