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

Gold Member

## Summary:

I need help in order to demonstrate a formal and rigorous proof that given a real function f is continuous that $f^+$ is also continuous ... ...

## Main Question or Discussion Point

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 $f$ is continuous then $f^+ = \text{max} (f, 0)$ is continuous ...

My thoughts are as follows:

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

Help will be much appreciated ...

Peter

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fresh_42
Mentor
Hi Peter,

you got the right point where it might be a problem, namely when $f$ changes to $0$. Now a proof depends on the definition you use for a continuous function. You used "preimages of open sets are open" recently. So take an open set around such a point where $f$ crosses the $x-$axis and determine its preimage.

Math_QED
Homework Helper
2019 Award
A more general thing is true. If $f,g$ are continuous functions, then so is $f\lor g=\max\{f,g\}$.

The easiest way to see this is to note that

$$f\lor g=\frac{1}{2}(f+g+|f-g|)$$

Gold Member
My thanks to Math_QED and fresh_42 for their help ...

Following the advice of fresh_42 my proof of the case where $f$ crosses the axis proceeds as follows:

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

... see Figure 1 below ...

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

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

so ...

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

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

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

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

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

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

The proof is similar if $f$ first crosses the x-axis from below ...

Is that correct?

Peter

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fresh_42
Mentor
You have written $f$, but we are interested in $f^+$. And the ordering is deliberate. Who guarantees you that $f(]x_1,x_2[)=]-y_1,y_1[$? What we want to know is, how the preimage of $(f^{+})^{-1}(]y_1,y_2[)$ looks like.

We start with an open interval $V\subseteq \mathbb{R}$. This is sufficient as any open set is a union of open intervals, and $(f^{+})^{-1}(\cup_\iota U_\iota) = \cup_\iota (f^{+})^{-1}(U_\iota)$ which is open if the individual sets are. Hence $V=]y_1,y_2[$. If $0\neq V$, then $f^+\equiv f$ or $f^+\equiv 0$ which are both continuous and we are done.

Thus we are left with $0\in V = ]y_1,y_2[$. What is $(f^{+})^{-1}(V)$ and how can we find out?

Gold Member
You have written $f$, but we are interested in $f^+$. And the ordering is deliberate. Who guarantees you that $f(]x_1,x_2[)=]-y_1,y_1[$? What we want to know is, how the preimage of $(f^{+})^{-1}(]y_1,y_2[)$ looks like.

We start with an open interval $V\subseteq \mathbb{R}$. This is sufficient as any open set is a union of open intervals, and $(f^{+})^{-1}(\cup_\iota U_\iota) = \cup_\iota (f^{+})^{-1}(U_\iota)$ which is open if the individual sets are. Hence $V=]y_1,y_2[$. If $0\neq V$, then $f^+\equiv f$ or $f^+\equiv 0$ which are both continuous and we are done.

Thus we are left with $0\in V = ]y_1,y_2[$. What is $(f^{+})^{-1}(V)$ and how can we find out?

Oh ... that was a typo ... made a mistake copying my notes without thinking ... have edited it now ...

Is the post still in error ...?

... will now reflect on what you have written ...

Peter

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fresh_42
Mentor
I do not see where you get this very specific form of $V$ from. We have to show that $(f^+)^{-1}(V)$ is open for any open set $V\subseteq \mathbb{R}$. See my previous post for why we can assume that $V$ is an interval.

Now if $0\notin V$ then $(f^+)^{-1}(V)$ is either empty or $(f^+)^{-1}(V)=f^{-1}(V)$, which both are open sets. Thus we may assume that $0\in V$ and we have to determine $(f^+)^{-1}(V)$ for this case. Yes, $V$ looks like $V=(y_1,0)\cup \{0\} \cup (0,y_2)$ in this case, but we cannot know what $f^+$ does there, so we cannot write $y_1=f^+(a_1)\, , \,y_2=f^+(a_2)\,.$ Maybe $f^+(x)=f(x)> 1000$ and we have chosen $y_2=1$. The task is: Show that $(f^+)^{-1}((y_1,0)\cup \{0\} \cup (0,y_2))$ is open for any pair $y_1<0<y_2\,.$

If we want to consider continuity especially at $x=c$ where $f$ crosses the $x-$axis, then you should use the analytical $\varepsilon-\delta$ definition of continuity, not the topological. If you go by open sets, then any open set $V$ is possible.

You can still use @Math_QED 's approach which uses the theorem: certain combinations of continuous functions are again continuous.

Gold Member
I do not see where you get this very specific form of $V$ from. We have to show that $(f^+)^{-1}(V)$ is open for any open set $V\subseteq \mathbb{R}$. See my previous post for why we can assume that $V$ is an interval.

Now if $0\notin V$ then $(f^+)^{-1}(V)$ is either empty or $(f^+)^{-1}(V)=f^{-1}(V)$, which both are open sets. Thus we may assume that $0\in V$ and we have to determine $(f^+)^{-1}(V)$ for this case. Yes, $V$ looks like $V=(y_1,0)\cup \{0\} \cup (0,y_2)$ in this case, but we cannot know what $f^+$ does there, so we cannot write $y_1=f^+(a_1)\, , \,y_2=f^+(a_2)\,.$ Maybe $f^+(x)=f(x)> 1000$ and we have chosen $y_2=1$. The task is: Show that $(f^+)^{-1}((y_1,0)\cup \{0\} \cup (0,y_2))$ is open for any pair $y_1<0<y_2\,.$

If we want to consider continuity especially at $x=c$ where $f$ crosses the $x-$axis, then you should use the analytical $\varepsilon-\delta$ definition of continuity, not the topological. If you go by open sets, then any open set $V$ is possible.

You can still use @Math_QED 's approach which uses the theorem: certain combinations of continuous functions are again continuous.

Hi fresh_42 ...

You write:

" ... ... I do not see where you get this very specific form of $V$ from. ... ... "

I'll try to explain ...

First, the context of my attempt at a proof was to show that $f^+$ was continuous at a point $c_1$ where $f$ crossed the x-axis from above ... that is the situation where point $c_1 \in \mathbb{R}$ is such that for $x \lt c_1$, $f(x) = f^+(x)$ is positive and for $x \geq c_1$, $f^+(x) = 0$ ... but I also considered f to be such that for some point $c_2 \gt c_1$ f crosses the x-axis from below ... ... that is for some point $c_2 \gt c_1$ we have that $f^+(x) = 0$ for $x \leq c_2$ and $f(x) = f^+(x)$ is positive for $x \gt c_2$ ... ...

The aim was to generalise from this specific form of $f$ and $f^+$ ... ...

Now for this proof I am following Andrew Browder's book "Mathematical Analysis: An Introduction" ...

Browder defines a neighbourhood of a point $a \in \mathbb{R}$ as follows:

and I also followed Browder Proposition 3.9 (d) winch reads as follows:

So following Definition 3.3 and Proposition 3.9 I formed a neighbourhood $V$ of $f^+(c_1)$ as

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

so ...

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

I hope that makes sense...

Peter

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fresh_42
Mentor
Definition 3.3. defines a neighborhood and the subset topology. I wanted to use definition (d) (or (c)) of proposition 3.9. The crucial point is in both cases: for every neighborhood $V$. So we start with an arbitrary neighborhood of $0\in V$ since $f(c)=0$. We may assume $0\in V=(y_1,y_2)$. We may not assume $y_i=f^+(a_i)$ since this wouldn't be an arbitrary $V$.

Take your picture, rotate it by $90°$ to the left

and calculate $(f^+)^{-1}(V)$ for various possibilities of $f^+$.

Gold Member
Definition 3.3. defines a neighborhood and the subset topology. I wanted to use definition (d) (or (c)) of proposition 3.9. The crucial point is in both cases: for every neighborhood $V$. So we start with an arbitrary neighborhood of $0\in V$ since $f(c)=0$. We may assume $0\in V=(y_1,y_2)$. We may not assume $y_i=f^+(a_i)$ since this wouldn't be an arbitrary $V$.

Take your picture, rotate it by $90°$ to the left
View attachment 255805
and calculate $(f^+)^{-1}(V)$ for various possibilities of $f^+$.

View attachment 255807

I have started to examine some possibilities for $f^+$ ... calculating $(f^+)^{-1} (V)$ for each ...

Examples are as follows:

For $f^+$ as shown in Figure 2 we have:

$V= \{ y \ : \ y_2 \lt y \lt y_1 \}$

so that

$(f^+)^{-1} (V) = \{ x \ : \ x_1 \lt x \lt x_2 \}$ ... ... ... ... ... which is an open set ...

For a second example consider $f^+$ as shown in Figure 3 below ... ...

For $f^+$ as shown in Figure 3 we have:

$V= \{ y \ : \ y_2 \lt y \lt y_1 \}$

so that

$(f^+)^{-1} (V) = \{ x \ : \ - \infty \lt x \lt x_2 \}$ ... ... ... ... ... which is an open set ...

For a third example consider $f^+$ as shown in Figure 4 below ... ...

For $f^+$ as shown in Figure 4 we have:

$V= \{ y \ : \ y_2 \lt y \lt y_1 \}$

so that

$(f^+)^{-1} (V) = \{ x \ : \ - \infty \lt x \lt x_1 \} \cup \{ x \ : \ x_2 \lt x \lt x_3 \}$ ... ... ... ... ... which is an open set ...

I have two questions ....

1. Is the above analysis correct as far as it goes ...?

2. If it is correct ... how do I use the analysis to organize and construct a general proof ... ...

Hope that you can help further ...

Peter

fresh_42
Mentor
Yes, this looks ok so far. But listing all possibilities is a bit inconvenient.

We can assume that $f^+(\mathbb{R})\cap V \neq \emptyset$ since otherwise $(f^+)^{-1}(V)=\emptyset$ which is open and we are done. We can also assume that $f(c)= 0\in V$ for otherwise we have again $(f^+)^{-1}(V)=\emptyset$ or $(f^+)^{-1}(V)=f^{-1}(V)$ which is open by continuity of $f$.

Our $V$ looks like $V=(y_1,0) \cup \{0\} \cup (0,y_2)$ with $y_1<0<y_2$ so $(f^+)^{-1}(V)=(f^+)^{-1}((y_1,0))\cup (f^+)^{-1}(\{0\}) \cup (f^+)^{-1}((0,y_2)=\emptyset \cup (f^+)^{-1}([0,y_2))=(f^+)^{-1}([0,y_2))$.

Now we only have to bother, whether $y_2\in f(\mathbb{R})$ or not. So what is $(f^+)^{-1}([0,y_2))$ in these two cases?
a.) $y_2=f(x_2)$ for some $x_2\in \mathbb{R}$
b.) $y_2 > f(\mathbb{R})$

Gold Member
Yes, this looks ok so far. But listing all possibilities is a bit inconvenient.

We can assume that $f^+(\mathbb{R})\cap V \neq \emptyset$ since otherwise $(f^+)^{-1}(V)=\emptyset$ which is open and we are done. We can also assume that $f(c)= 0\in V$ for otherwise we have again $(f^+)^{-1}(V)=\emptyset$ or $(f^+)^{-1}(V)=f^{-1}(V)$ which is open by continuity of $f$.

Our $V$ looks like $V=(y_1,0) \cup \{0\} \cup (0,y_2)$ with $y_1<0<y_2$ so $(f^+)^{-1}(V)=(f^+)^{-1}((y_1,0))\cup (f^+)^{-1}(\{0\}) \cup (f^+)^{-1}((0,y_2)=\emptyset \cup (f^+)^{-1}([0,y_2))=(f^+)^{-1}([0,y_2))$.

Now we only have to bother, whether $y_2\in f(\mathbb{R})$ or not. So what is $(f^+)^{-1}([0,y_2))$ in these two cases?
a.) $y_2=f(x_2)$ for some $x_2\in \mathbb{R}$
b.) $y_2 > f(\mathbb{R})$

Thanks again for your help, fresh_42 ...

You write:

" ... ...
So what is $(f^+)^{-1}([0,y_2))$ in these two cases?
a.) $y_2=f(x_2)$ for some $x_2\in \mathbb{R}$
b.) $y_2 > f(\mathbb{R})$ ... ... "

Based on what was indicated by the examples I considered ...

... if $y_2=f(x)$ for some $x \in \mathbb{R}$

... then ...

$(f^+)^{-1} (V) = \{ x \ : \ x_1 \lt x \lt x_2 \} \cup \{ x \ : \ x_3 \lt x \lt x_4 \} \cup \ldots \cup \{ x \ : \ x_{n-1} \lt x \lt x_n \}$

where $x_1$ and/or $x_n$ may have to be replaced by $- \infty$ and $+ \infty$ respectively ... and the number of unions may be countably infinite ...

In all these cases $(f^+)^{-1} (V)$ is an open set ...

... ...

If $y_2 > f(\mathbb{R})$ ...

... then ...

$(f^+)^{-1} (V) = \emptyset$ ... ... and $\emptyset$ is an open set ...

Is the above correct?

Peter

*** EDIT ***

Correction to the following:

" ... ...

If $y_2 > f(\mathbb{R})$ ...

... then ...

$(f^+)^{-1} (V) = \emptyset$ ... ... and $\emptyset$ is an open set ... ... "

If $y_2 > f(\mathbb{R})$ ...

... then ...

$(f^+)^{-1} (V) = \mathbb{R}$ ... ... and $\mathbb{R}$ is an open set ...

Hope that part of my answer is now correct ...

Also suspect that I was in error in not dealing with 0 properly in the expression $(f^+)^{-1}([0,y_2))$

Peter

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fresh_42
Mentor
No, it's not. Don't make it too complicated with the many subintervals. We do not need to and cannot know the details of $f$. What we need is $(f^+)^{-1}(V)=(f^+)^{-1}([0,y_2))$.

We also said that we may assume that f touches the $x-$axis, so we have either $(f^+)^{-1}([f^+(c),f^+(x)))$ or $(f^+)^{-1}([0,y_2))$ where $y_2> f^+(x)$ for all $x$.

Now forget for a moment the specific situation. Let's assume we have a function $g: \mathbb{R} \longrightarrow [0,\infty)$. What is $g^{-1}([g(0),g(x)))$ and what is $g^{-1}([0,C))$ where $\operatorname{im}(g)=g(\mathbb{R})\subseteq [0,C)$ since $C$ is bigger than any function value?

What are the pre-images in these cases? Solve it for $g$ and replace $g$ by $f^+$ afterwards.

Gold Member
No, it's not. Don't make it too complicated with the many subintervals. We do not need to and cannot know the details of ff. What we need is (f+)−1(V)=(f+)−1([0,y2))(f+)−1(V)=(f+)−1([0,y2)).

We also said that we may assume that f touches the x−x−axis, so we have either (f+)−1([f+(c),f+(x)))(f+)−1([f+(c),f+(x))) or (f+)−1([0,y2))(f+)−1([0,y2)) where y2>f+(x)y2>f+(x) for all xx.

Now forget for a moment the specific situation. Let's assume we have a function g:R⟶[0,∞)g:R⟶[0,∞). What is g−1([g(0),g(x)))g−1([g(0),g(x))) and what is g−1([0,C))g−1([0,C)) where im(g)=g(R)⊆[0,C)im⁡(g)=g(R)⊆[0,C) since CC is bigger than any function value?

What are the pre-images in these cases? Solve it for gg and replace gg by f+f+ afterwards.

" ... ... Let's assume we have a function g:R⟶[0,∞). What is g−1([g(0),g(x))) and what is g−1([0,C)) where im(g)=g(R)⊆[0,C) since C is bigger than any function value? ... ... "

Consider g−1([g(0),g(x))) given g as follows:

We have g−1([g(0),g(x)))=[0,x) ... ...

But not quite sure why we are determining this pre-image ...

Now consider g−1([0,C)) where im(g)=g(R)⊆[0,C) since C is bigger than any function value ... where we are given g as in Figure 6 below

We find g−1([0,C))=R

Can you help further ...

Peter

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fresh_42
Mentor
$g^{-1}([g(0),x)) =\{\,u\in \mathbb{R}\,|\,0\leq g(u)< g(x)\,\} = \{\,u\in \mathbb{R}\,|\, g(u)< g(x)\,\}$

We need the definition of $f^+$ faster than I thought, sorry. I have been too concentrated on how to get rid of the boundary $g(0)$ on the left.

What we now have is $(f^+)^{-1}([(f^+)(0),x)) =\{\,u \in \mathbb{R}\,|\,(f^+)(u)<(f^+)(x)\,\}$ and the next step is $\{\,u \in \mathbb{R}\,|\,(f^+)(u)<(f^+)(x)\,\} = \{\,u \in \mathbb{R}\,|\,f(u)<f(x)\,\}=f^{-1}(\;(-\infty,f(x)\;)$ which is open because $f$ is continuous.

Gold Member
$g^{-1}([g(0),x)) =\{\,u\in \mathbb{R}\,|\,0\leq g(u)< g(x)\,\} = \{\,u\in \mathbb{R}\,|\, g(u)< g(x)\,\}$

We need the definition of $f^+$ faster than I thought, sorry. I have been too concentrated on how to get rid of the boundary $g(0)$ on the left.

What we now have is $(f^+)^{-1}([(f^+)(0),x)) =\{\,u \in \mathbb{R}\,|\,(f^+)(u)<(f^+)(x)\,\}$ and the next step is $\{\,u \in \mathbb{R}\,|\,(f^+)(u)<(f^+)(x)\,\} = \{\,u \in \mathbb{R}\,|\,f(u)<f(x)\,\}=f^{-1}(\;(-\infty,f(x)\;)$ which is open because $f$ is continuous.

Hi fresh_42 ...

Sorry to be slow ... but what is $x$ in the expression $g^{-1}([g(0),x))$ ... ?

Further ... could you please explain why/how ...

$g^{-1}([g(0),x)) =\{\,u\in \mathbb{R}\,|\,0\leq g(u)< g(x)\,\} = \{\,u\in \mathbb{R}\,|\, g(u)< g(x)\,\}$

Again ... sorry if I am asking the obvious ...

*** EDIT *** ... oh! presumably its just the definition of pre-image ...?

Peter

fresh_42
Mentor
$x$ was a very bad choice. It forced me to switch to $u$ as variable name. I made this mistake in post #13.
Better would have been the cases: $[0,y_2)$ with $(f^+)(c)=0, (f^+)(x_2)=y_2$ and $[0,C)$ with $f^+(x)<C \;\forall\,x\in \mathbb{R}$.

It was meant as a point $(x,y)=(x_2,y_2=g(x_2))$ with some given image point on the right.

Edit: $x_2$ has to be chosen as $x_2=\sup\{\,x\in \mathbb{R}\,|\,f(x)=y_2\,\}.$

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Gold Member
$x$ was a very bad choice. It forced me to switch to $u$ as variable name. I made this mistake in post #13.
Better would have been the cases: $[0,y_2)$ with $(f^+)(c)=0, (f^+)(x_2)=y_2$ and $[0,C)$ with $f^+(x)<C \;\forall\,x\in \mathbb{R}$.

It was meant as a point $(x,y)=(x_2,y_2=g(x_2))$ with some given image point on the right.

OK ... understand ...

Thanks once again for all your help ...

Peter

Gold Member
Hi fresh_42 ...

I thought I'd just (for completeness) set out the proof of the continuity of $f^+$ in terms of the variables we were talking about early on ...

Consider Case 1: $y_2 = f^+ (x_2)$ for some $x_2 \in \mathbb{R}$ ... ...

See Figure 7 below ...

We have $y_2 = f^+ (x_2) = f(x_2)$ ... and ... $f^+(c) = f(c) = 0$ ...

We are required to calculate $(f^+)^{-1} ( [0, y_2) )$ ...

We have ...

$(f^+)^{-1} ( [0, y_2) ) = \{ x\ : \ 0 \leq f^+ (x) \lt f^+ (x_2) \}$

... so ... $(f^+)^{-1} ( [0, y_2) ) = \{ x\ : \ f^+ (x) \lt f^+ (x_2) \}$

... therefore ... $(f^+)^{-1} ( [0, y_2) ) = \{ x\ : \ f(x) \lt f(x_2) \}$

... and therefore ... $(f^+)^{-1} ( [0, y_2) ) = f^{-1}( ( - \infty , f(x_2) ) )$

... and $f^{-1}( ( - \infty , f(x_2) ) )$ is an open set since $f$ is continuous ...

Now consider Case 2 where $y_2 \gt f( \mathbb{R} )$ ... see Figure 8 below ...

In Case 2 we have $(f^+)^{-1} ( [0, y_2) ) = \mathbb{R}$ ...

... and $\mathbb{R}$ is an open set ...

Hopefully the above is correct...

Peter