If function f is increasing, does f^2 increase?

I have a question, take f:R-> R is a real function. If f increases, then f^2 increases? Does this statement hold for the converse and contrapositive?

Can you think of any examples where it's not true?

The statement holds if and only if the contrapositive holds, so you only need to worry about the converse.

so f(x1) < f(x2). Then f^2(x1) < f^2(x2)?

That's not true. For example, $-3 < -2$ but $(-3)^2=9>4=(-2)^2$. This example should give you a hint at some sort of counterexample.


The contrapositive wouldn't say anything about it being decreasing. It would be about it "not being increasing". A function which isn't increasing isn't necessarily decreasing (e.g. $y=\sin x$ is neither). These are the statements you have:

Statement: if f is increasing then f² is increasing
Contrapositive: if f² isn't increasing then f isn't increasing
Converse: if f² is increasing then f is increasing

Ah thanks, that cleared up things for me. I was looking around the internet and some put $f^2 = f(f(x))$. I thought $f^2$ was $f$ squared, but which is right?

Personally my preference is to write $f^2 (x) = f(f(x))$ and $f(x)^2$ for, well, $f(x)^2$. I try to avoid writing plain $f^2$, unless putting in all the implied arguments gets messy. You can sometimes figure out which is meant by looking at the type of the function - if it's, say, $\mathbb{R}^3 \to \mathbb{R}$, then $f \circ f$ doesn't even make sense, so $f^2$ necessarily means $x \mapsto f(x)^2$. If it's $\mathbb{R}^3 \to \mathbb{R}^3$, then $f \circ f$ makes sense, but what does $f(x)^2$ mean? (Yes, usually it means the scalar product $f(x) \cdot f(x)$. But you may not be working in a context where that makes sense.)

What if it's $f:\mathbb{R} \to \mathbb{R}$? Should I use both $f^2$ as $f$ squared, and $f(f(x))$? $f(x)^2$ doesn't make sense in my opinion.

I'm thinking if $f^2$ is indeed $f(f(x))$, then what I was thinking maybe I could do:

Take $f(x)=ax+b$ as our increasing function,defined on $f:\mathbb{R} \to \mathbb{R}$ , then $g(x)=f(f(x))= a(ax +b) +b$.

Hence $g(x)=f(f(x))= a^2x +ab +b$.

Differentiate $g(x)$ to get $g'(x) =a^2$, which is always going to be positive and increasing since $g'(x)=a^2 >0$ for all $a$ which is a real number.

So what is it going to be for the converse and contrapositive statements?

For what it's worth, I'd write $(f \circ f)(x)$ for $f(f(x))$. If you had $n$ compositions, say $f(f(\cdots(f(x))\cdots)$ then I'd write $f^{\circ n}(x)$; and for $(f(x))^n$ I'd write $f(x)^n$.

Interesting. I've considered doing something like that before, but the purist in me says that $f^2$ for the map $x \mapsto f(x)^2$ is just plain wrong and we should not need to disambiguate. (Along the same lines, writing out the circles for composition should be unnecessary, yet my aesthetic preference is to write them out. I should probably resolve this contradiction at some point...)

Just got told this morning that for this question, $f^2(x)$ is $(f(x))^2$. Now it's half the trouble for me.

So far the statement doesn't hold eg:.

Now I working towards the converse of the statement. If$f^2$ is increasing, then $f$ increasing.

What I got so far is let $g(x)=f^2(x)=x^(2n+1)$ for $n=0,1,2,...$, which is obviously and increasing function since $g'(x)=(2n+1)x^(2n)>0$ for all $x$.

Then $f(x)=\sqrt x^(2n+1) =x^(n+1/2)$. Differentiate $f(x)$, I get$f'(x)=(n+1/2)x^(n-1/2)$.

But then there is a problem. If I take n=0, then $f'(x)=1/(2\sqrt x)>0$. I cannot take -ve x values, only x.

My friend said she got converse as false and contrapositive statement as true. Hmm...

For the converse counterexample, let $f(x)=x^2$, then it's increasing for $f(x)=x$ but decreasing for $f(x)=-x$. Then the converse doesn't hold.

I'm a bit muddled with which contrapositive and converse we're talking about here (and also I'm guessing you meant $f^2(x)=x^2$?).

Basically we have 4 statements:
1. f is increasing => f² is increasing
2. f² is not increasing => f is not increasing
3. f² is increasing => f is increasing
4. f isn't increasing => f² isn't increasing

(1) and (2) are always either both true or both false, since (2) is the contrapositive of (1) and vice versa. Also, (3) and (4) are always either both true or false, for the same reason. (3) is the converse of (1). We've shown that (1) is false earlier in the thread, and your example here has shown that (3) is false (assuming we're only looking at the region $x>0$, since x² is decreasing on $x<0$!) and so (2) and (4) are both necessarily false.

The contrapositive for the converse (i.e. (4)) is "if f isn't increasing then f² isn't increasing". Well, as you've just shown, $f(x)=-x$ isn't increasing on the interval $x \ge 0$, but $f^2(x)=x^2$ is increasing on this interval, so this is false.

Ah, I was meant to say $f^2(x)=x^2$. This is now corrected.

So to summarise what you are saying:
1. If $f$ is increasing, then $f^2$ is increasing. False as given in a previous example.
2. If $f^2$ is increasing, then $f$ is increasing. False as given in a previous example.
3. If $f^2$ is not increasing, then $f$ is not increasing. False since statement 1 is false. (counterexample would be similar to counterexample in 1?)

Does it mean all are false statements?