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This is question from Spivak's Calculus. Question statement (paraphrased): For any even function $f$ there is infinite number of functions $g$ such that $f(x) = g(|x|)$
I have made attempt at proof, here is my work. My main concern is whether my proof that there are infinite such functions is correct.

Let $g$ be a function such that for any positive $x$, $f(x) = g(x)$ and let $f$ be some even function. We will show that above equality then holds even when $x$ is negative: $f(-x) = g(|-x|) \Leftrightarrow f(x) = g(x)$, which is true by definition of these functions. It now remains to show that there is infinite number of those functions. I understand that is so because $g$ can be defined as one whishes for negative numbers and it won't change equality. But I cannot think of formal proof. I'm thinking about assigning values to functions $g$ in a way that I define $g_{1}(-1) = k$ and then making infinite number of them by induction: for any function $g_{n}(x) = k$, if x is negative, we define $g_{n+1}(x-1) = k$. Then by the fact that there is infinite natural numbers, and for every natural number there is going to be unique function, we can conclude there is infinite number of functions.

Is the concept of induction applicable here or is it not?

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Induction in terms of "one-at-a-time" doesn't work for real numbers (but see However, it seems a bit silly; you already know that it doesn't matter how you define $g$ on the negative numbers, as long as $g(x)=f(x)$ for $x\geq 0$. So, why not define $g_k(x) = k$ for all $x\lt 0$, and $g_k(x)=f(x)$ for all $x\geq 0$? Different values of $k$ give different $g$s, so it's just a question of how many different $k$ you can pick. I think you'll have enough choices there... – Arturo Magidin Feb 6 '11 at 0:55
Everything here up to "But I cannot think of a formal proof" is correct. However, I found your description of your family of functions a little confusing (partly because you don't say much about $g_1$ other than the value $g_1(-1)$). Maybe you can find an easier choice for your family of functions $g_n$. (After all, there are a lot choices!) – Matt E Feb 6 '11 at 0:57
@Arturo: To be honest I don't see how I used induction on real numbers. I used induction on numbering of my functions, such that every function with different numbering is different. But I agree with other part of the post. Thank you. – user5501 Feb 6 '11 at 1:01
@Matt: My family is such that it retains all of the properties of defined $g$ and adds that value. – user5501 Feb 6 '11 at 1:03
@LovreP: I think I figured it out. Your $g_1$ is defined only on nonnegative reals and $-1$; then your $g_2$ is defined only on nonnegative reals and $-1$, and $-2$. And so on? Then it doesn't work. Your functions $g$ have to be defined on all real numbers. So you need to say what the value of $g_1$ is at, say, $-0.5$, and $-\pi$, and all negative real numbers, not just $-1$. Otherwise, you don't get $f(x) = g(|x|)$. – Arturo Magidin Feb 6 '11 at 1:06
up vote 3 down vote accepted

Okay, from the comments I was able to guess at what you were trying to do.

You are correct that the key is that as long as $g(x)=f(x)$ for $x\geq 0$, everything will work out, so you only need to worry about defining $g$ on the negative numbers.

You attempted to do so by letting $g_1$ be a function that is defined only on the nonnegative numbers and $-1$, and setting $g_1(-1)=k$. Then $g_2$ would be an extension of $g_1$, which is also defined as $k$ at $-2$; and so on. In general, $g_{n}$ would be define on $\{-n,-n+1,\ldots,-1\}\cup[0,\infty)$, by $g(x) = f(x)$ if $x\geq 0$ and $g(x)=k$ if $x\lt n$.

As I noted in the comments, your formulas didn't really say that; instead, they only specified $g_1$ at $-1$, and then said, for example, that $g_2(x-1)=k$. That is at best confusing. What is $g_2(-0.5)$? According to this, I have to think of $-0.5$ as $0.5 - 1$, and then it's $k$, and ... Well, a bit of a confusing issue arises...

In any case, whether this works as an answer or not depends on whether you are assuming that your functions need to be defined on the same set or not. Normally, we would be looking for functions $g$ such that $f(x)=g(|x|)$ and we want both $f$ and $g$ to have the same domain. Remember that two functions are equal if and only if they have the same domain, the same codomain, and the same value at every element of the domain. So even if you could have set up the induction properly to get the functions you wanted (or if you wanted to define $g_1$ on $[-1,0)$, then $g_2$ extended to all of $[-2,0)$, and so on sothat $g_n$ was defined on $[-n,\infty)$) it still would not give a good answer to the problem because of the restrictive domains of your $g$s. (You are really composing the absolute value with $g$; I think Spivak wants you to play with functions that are defined everywhere here, rather than on artificial domains; I could be wrong, though).

If you do not require your functions $g$ to have the same domain as $f$, then your intended answer would also work; the functions $g_n$ are different because they have different domains, even though they all have the same values where their domains agree.

I think, though, that the intended answer relies instead in defining $g_n(x)$ for $x\lt 0$ as different things for different $n$. For example, you could set $$g_n(x) = \left\{\begin{array}{ll} f(x) &\mbox{if $x\geq 0$,}\\ n &\mbox{if $x\lt 0$;} \end{array}\right.$$ and this would work. There is no need to state it as induction, because the values depend only on the labels, and you simply have infinitely many distinct labels to choose from.

That said: yes, you can use induction to define a (countably infinite) series of functions. If you wanted to do that here, you could express it explicitly stating what $g_1$ is (giving its domain clearly; if all you say is $g(-1)=k$, then you are telling us the value at $-1$, but not saying anything about values elsewhere, not even that it is not defined there). Then saying that assuming you have defined $g_n$ with a domain of, say, $[-n,\infty)$, you define $g_{n+1}$ on $[-n-1,\infty)$ by $$g_{n+1}(x) = \left\{\begin{array}{ll} g_n(x) &\mbox{if $x\in[-n,\infty)$.}\\ \mbox{whatever} &\mbox{if $x\in [-n-1,-n)$;} \end{array}\right.$$ This would indeed give an inductive definition for your $g_n$, defined on $[-n,\infty)$, each extending the previous one.

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Thank you a lot for your time, I learned a lot reading your posts. – user5501 Feb 6 '11 at 1:44
I think that you meant $[-2,0)$ at the fifth paragraph, and $[-n-1,-n)$ in your definition of $g_{n+1}$. Please correct me if I'm wrong in either case. – user5501 Feb 6 '11 at 1:45
@LovreP: Yes on both; thanks. – Arturo Magidin Feb 6 '11 at 1:49

It is enough to observe that $g$ can have any value whatsoever for $x < 0$. Since there is an infinite number of values for each $f(x)$ for $x < 0$, we're done.

(you really don't need to do it any more difficult than that.)

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Close. It is not that there are many values of x<0, but many possible values of g(x) for x<0. Imagine that something in the problem forced g(x) to be 0 for x<0. You still have lots of x, but only one g. – Ross Millikan Feb 6 '11 at 1:00
@Frederik: Actually, you are being somewhat misleading. It's not that there are an infinite number of $x$ that are less than 0 by itself, it's that you have a choice about where to send them. The result would still be true if your function was only defined on $\{-1,0,1\}$. (The only way you cannot is if you have a finite number of negative numbers, and a finite number of possible images...) – Arturo Magidin Feb 6 '11 at 1:01
I'm being silly of course. Thanks for the correction. – Fredrik Meyer Feb 6 '11 at 1:04
I see. Thank you for your post. – user5501 Feb 6 '11 at 1:04

You could use induction but it's really not necessary. Just define $g_k(x) = k$ when $x$ is negative. Then you have a infinite family of functions that are different.

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Thanks for your post. – user5501 Feb 6 '11 at 1:52

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