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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 http://math.stackexchange.com/questions/4202/induction-on-real-numbers). 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 06 '11 at 00:55
  • 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 06 '11 at 00:57
  • @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. –  Feb 06 '11 at 01:01
  • @Matt: My family is such that it retains all of the properties of defined $g$ and adds that value. –  Feb 06 '11 at 01:03
  • It's unclear what you are doing; tell me, what is your $g_2$? The problem is that you are also "moving the argument" (by writing $x-1$), which is why it seemed like you were trying to do induction on the argument (as well as on the index). – Arturo Magidin Feb 06 '11 at 01:03
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    @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 06 '11 at 01:06
  • @Arturo: $g_{2}$ is given inductively from $g_{1}$. Since $g_{1}(-1) = k$, then $g_{1+1}(-1-1)=k$. Thank you for your post. –  Feb 06 '11 at 01:09
  • @Arturo: I see. Thank you, I now see flaw in my reasoning. –  Feb 06 '11 at 01:10
  • @Arturo: Can you please post that as an answer so I can accept it? While other answers have given me means of proving that statement, you are the only one that addressed whether my induction argument fails. –  Feb 06 '11 at 01:12
  • @Arturo:That post was written before I have seen your clarification. Sorry for that, I also made that typo with $g_{-1}$. –  Feb 06 '11 at 01:13
  • @LovreP: Please note that your formulas did not in fact define $g_2$ inductively; perhaps what I wrote is what you meant, but I'm afraid that it is not at all what you wrote. What you wrote does not define $g_{n+1}$ the way I described (which I was just guessing about). – Arturo Magidin Feb 06 '11 at 01:13
  • @LovreP: Which post as an answer? First one, or "I think I figured it out..." or a combination thereof? – Arturo Magidin Feb 06 '11 at 01:14
  • @Arturo: I see. Yes, that was what I meant. I'm sorry, I thought that was proper way to define it. –  Feb 06 '11 at 01:16
  • @Arturo: "I think I figured it out" or combination. Thank you for your time. –  Feb 06 '11 at 01:17
  • @Arturo: I'm really sorry, but I'm not seeing it again. How would I not get that equality, if $g(x)$ where $x$ is negative never gets evaluated in it because of absolute value? –  Feb 06 '11 at 01:19
  • @LovreP: Because two functions are equal if and only if they have the same domain, the same codomain, and the same values at every element of the domain. You want your functions $g$ to have the same domain as $f$, even if you only evaluate them at nonnegative reals. – Arturo Magidin Feb 06 '11 at 01:26
  • @Arturo: I see, but question doesn't ask for the functions to be equal, just to for equality to hold. If we go by that, then how there is going to be infinite number of functions? That is, then $g$ has to be defined same for negative $x$ as $f$ because of your requirement "the same values at every element of the domain". –  Feb 06 '11 at 01:30
  • @No, it doesn't have to be the same as $f$ on the negative numbers, just be defined on negative numbers. That said, your intended answer could be argued to answer the question correctly, if you are not assuming that your functions are defined on all real numbers. – Arturo Magidin Feb 06 '11 at 01:33
  • @Arturo: I see. Thank you for your time. –  Feb 06 '11 at 01:34
  • @LovreP: I've edited my answer to account for that. Your intended functions would indeed be pairwise distinct, so I think that, absent context, they can be taken as correct answers. I've also given a way in which you could have defined them inductively (though on intervals, rather than just on negative integesr). – Arturo Magidin Feb 06 '11 at 01:38

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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.

Arturo Magidin
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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.)

Fredrik Meyer
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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 06 '11 at 01:00
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    @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 06 '11 at 01:01
  • I'm being silly of course. Thanks for the correction. – Fredrik Meyer Feb 06 '11 at 01:04
  • I see. Thank you for your post. –  Feb 06 '11 at 01:04
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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.