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

1. 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

2.

3. 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.

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 x-axis that I am having trouble dealing with ...

Peter

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 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

6. 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)".