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Continuity of f^+ ... Browder Corollary 3.13

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,918
I am reading Andrew Browder's book: "Mathematical Analysis: An Introduction" ... ...

I am currently reading Chapter 3: Continuous Functions on Intervals and am currently focused on Section 3.1 Limits and Continuity ... ...

I need some help with the proof of Corollary 3.13 ...


Corollary 3.13 reads as follows:


Browder - Corollary 3.13 ... .png


Can someone help me to prove that if \(\displaystyle f\) is continuous then \(\displaystyle f^+ = \text{max} (f, 0)\) is continuous ...


My thoughts are as follows:


If \(\displaystyle c\) belongs to an interval where \(\displaystyle f\) is positive then \(\displaystyle f^+\) is continuous since \(\displaystyle f\) is continuous ... further, if \(\displaystyle c\) belongs to an interval where \(\displaystyle f\) is negative then \(\displaystyle f^+\) is continuous since \(\displaystyle g(x) = 0\) is continuous ... but how do we construct a proof for those points where \(\displaystyle f(x)\) crosses the \(\displaystyle x\)-axis ... ..




Help will be much appreciated ...

Peter
 
Last edited:

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
To prove [tex]f^+[/tex] is continuous at [tex]x_0[/tex] consider 3 cases:
1) [tex]f(x_0)> 0[/tex].
2) [tex]f(x_0)= 0[/tex].
3) [tex]f(x_0)< 0[/tex].

Since f is continuous, in case (1) there exist an interval around [tex]x_0[/tex] such that f(x)> 0
and [tex]f^+(x)= f(x)[/tex]
for all x in the interval

Since f is continuous, in case (1) there exist an interval around [tex]x_0[/tex] such that f(x)< 0 and [tex]f^+(x)= 0[/tex] for all x in the interval.
 

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,918
To prove [tex]f^+[/tex] is continuous at [tex]x_0[/tex] consider 3 cases:
1) [tex]f(x_0)> 0[/tex].
2) [tex]f(x_0)= 0[/tex].
3) [tex]f(x_0)< 0[/tex].

Since f is continuous, in case (1) there exist an interval around [tex]x_0[/tex] such that f(x)> 0
and [tex]f^+(x)= f(x)[/tex]
for all x in the interval

Since f is continuous, in case (1) there exist an interval around [tex]x_0[/tex] such that f(x)< 0 and [tex]f^+(x)= 0[/tex] for all x in the interval.


Thanks for the help ...

BUT ... you do not describe what to do in case 2 ... and as I indicated it is when f(x) = 0, specifically when f crosses the x-axis that I am having trouble dealing with ...

Peter
 

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,918
Thanks for the help ...

BUT ... you do not describe what to do in case 2 ... and as I indicated it is when f(x) = 0, specifically when f crosses the x-axis that I am having trouble dealing with ...

Peter



After reflecting on this problem for some time ... here is my proof for the situation where the point investigated is a point where \(\displaystyle f\) crosses the x-axis ...


I think it will suffice to prove that \(\displaystyle f^+\) is continuous for the case where a point \(\displaystyle c_1 \in \mathbb{R}\) is such that for \(\displaystyle x \lt c_1, \ f(x) = f^+(x)\) is positive and for \(\displaystyle x \gt c_1, \ f^+(x) = 0\) ... ... while for some point \(\displaystyle c_2 \gt c_1\) we have that \(\displaystyle f^+(x) = 0\) for \(\displaystyle x \lt c_2\) and \(\displaystyle f(x) = f^+(x)\) is positive for \(\displaystyle x \gt c_2\) ... ...


... see Figure 1 below ...




Figure 1 - Continuity of f+ ... .png





Now consider an (open) neighbourhood \(\displaystyle V\) of \(\displaystyle f^+(c_1)\) where...



\(\displaystyle V= \{ f^+(x) \ : \ -f^+(a_1) \lt f^+(c_1) \lt f^+(a_1)\) for some \(\displaystyle a_1 \in \mathbb{R} \}

\)

so ...


\(\displaystyle V= \{ f^+(x) \ : \ -f^+(a_1) \lt 0 \lt f^+(a_1)\) for some \(\displaystyle a_1 \in \mathbb{R} \}\) ...




Then ... (see Figure 1) ...



\(\displaystyle (f^+)^{ -1 } (V) = \{ a_1 \lt x \lt a_2 \}\) which is an open set as required ....



Further crossings of the x-axis by \(\displaystyle f\) just lead to further sets of the nature \(\displaystyle \{ a_{ n-1 } \lt x \lt a_n \}\) which are also open ... so ...




\(\displaystyle (f^+)^{ -1 } (V) = \{ a_1 \lt x \lt a_2 \} \cup \{ a_3 \lt x \lt a_4 \} \cup \ldots \cup \{ a_{n-1} \lt x \lt a_n \}\)



which being a union of open sets is also an open set ...




The proof is similar if \(\displaystyle f\) first crosses the x-axis from below ...







Is that correct?


Peter
 
Last edited:

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
To prove [tex]f^+[/tex] is continuous at [tex]x_0[/tex] consider 3 cases:
1) [tex]f(x_0)> 0[/tex].
2) [tex]f(x_0)= 0[/tex].
3) [tex]f(x_0)< 0[/tex].

Since f is continuous, in case (1) there exist an interval around [tex]x_0[/tex] such that f(x)> 0
and [tex]f^+(x)= f(x)[/tex]
for all x in the interval

Since f is continuous, in case (1) there exist an interval around [tex]x_0[/tex] such that f(x)< 0 and [tex]f^+(x)= 0[/tex] for all x in the interval.
TYPO: in the last sentence "case (1)" should have been "case (2)".