
MHB Master
#1
January 17th, 2020,
02:48
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:
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 by Peter; January 17th, 2020 at 05:00.

January 17th, 2020 02:48
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#2
January 17th, 2020,
08:21
To prove $ \displaystyle f^+$ is continuous at $ \displaystyle x_0$ consider 3 cases:
1) $ \displaystyle f(x_0)> 0$.
2) $ \displaystyle f(x_0)= 0$.
3) $ \displaystyle f(x_0)< 0$.
Since f is continuous, in case (1) there exist an interval around $ \displaystyle x_0$ such that f(x)> 0 and $ \displaystyle f^+(x)= f(x)$
for all x in the interval
Since f is continuous, in case (1) there exist an interval around $ \displaystyle x_0$ such that f(x)< 0 and $ \displaystyle f^+(x)= 0$ for all x in the interval.

MHB Master
#3
January 17th, 2020,
23:50
Thread Author
Originally Posted by
HallsofIvy
To prove $ \displaystyle f^+$ is continuous at $ \displaystyle x_0$ consider 3 cases:
1) $ \displaystyle f(x_0)> 0$.
2) $ \displaystyle f(x_0)= 0$.
3) $ \displaystyle f(x_0)< 0$.
Since f is continuous, in case (1) there exist an interval around $ \displaystyle x_0$ such that f(x)> 0
and $ \displaystyle f^+(x)= f(x)$
for all x in the interval
Since f is continuous, in case (1) there exist an interval around $ \displaystyle x_0$ such that f(x)< 0 and $ \displaystyle f^+(x)= 0$ 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 xaxis that I am having trouble dealing with ...
Peter

MHB Master
#4
January 18th, 2020,
02:34
Thread Author
Originally Posted by
Peter
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 xaxis 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 xaxis ...
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 ...
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 xaxis by $ \displaystyle f$ just lead to further sets of the nature $ \displaystyle \{ a_{ n1 } \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_{n1} \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 xaxis from below ...
Is that correct?
Peter
Last edited by Peter; January 18th, 2020 at 20:58.

#5
January 18th, 2020,
09:37
Originally Posted by
HallsofIvy
To prove $ \displaystyle f^+$ is continuous at $ \displaystyle x_0$ consider 3 cases:
1) $ \displaystyle f(x_0)> 0$.
2) $ \displaystyle f(x_0)= 0$.
3) $ \displaystyle f(x_0)< 0$.
Since f is continuous, in case (1) there exist an interval around $ \displaystyle x_0$ such that f(x)> 0
and $ \displaystyle f^+(x)= f(x)$
for all x in the interval
Since f is continuous, in case (1) there exist an interval around $ \displaystyle x_0$ such that f(x)< 0 and $ \displaystyle f^+(x)= 0$ for all x in the interval.
TYPO: in the last sentence "case (1)" should have been "case (2)".