# Continuity of f^+ ... Browder Corollary 3.13

#### Peter

##### Well-known member
MHB Site Helper
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 ...

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

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#### HallsofIvy

##### Well-known member
MHB Math Helper
To prove $$f^+$$ is continuous at $$x_0$$ consider 3 cases:
1) $$f(x_0)> 0$$.
2) $$f(x_0)= 0$$.
3) $$f(x_0)< 0$$.

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

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

#### Peter

##### Well-known member
MHB Site Helper
To prove $$f^+$$ is continuous at $$x_0$$ consider 3 cases:
1) $$f(x_0)> 0$$.
2) $$f(x_0)= 0$$.
3) $$f(x_0)< 0$$.

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

Since f is continuous, in case (1) there exist an interval around $$x_0$$ such that f(x)< 0 and $$f^+(x)= 0$$ 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 x-axis that I am having trouble dealing with ...

Peter

#### Peter

##### Well-known member
MHB Site Helper
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 ...

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

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#### HallsofIvy

##### Well-known member
MHB Math Helper
To prove $$f^+$$ is continuous at $$x_0$$ consider 3 cases:
1) $$f(x_0)> 0$$.
2) $$f(x_0)= 0$$.
3) $$f(x_0)< 0$$.

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

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