Uniqueness of Even and Odd Function Decomposition

[avec_holl]

Homework Statement

13. (a) Prove that any function $f$ with domain $\mathbb{R}$ can be written $f = O + E$ where $E$ is even and $O$ is odd.

(b) Prove that this way of writing $f$ is unique.

Homework Equations

N/A

The Attempt at a Solution

(a) Suppose that $f$ is some function with domain $\mathbb{R}$. This implies that for any arbitrary real number $a$, both $f(a)$ and $f(-a)$ exist. We can now define even and odd functions such that $f(a) = O(a) + E(a)$ and $f(-a) = E(a) - O(a)$. Since $f$ can be represented by the sum of even and odd functions at a single point, we can define $O$ and $E$ point-wise such that their sum is always equal to $f$.

I'm not sure if this is a reasonable argument, but if it is, how can I make it rigorous?


(b) Suppose not, then $f = O_1 + E_1 = O_2 + E_2$. This implies that $O_1 = (E_2 - E_1) + O_2$. However, since $O_1$ is necessarily odd and $E_2 - E_1$ is necessarily even, we have that $E_1 = E_2$ and consequently $O_1 = O_2$. Hence, this way of writing $f$ is unique.

I'm not sure if this is a reasonable argument either and any suggestions would be appreciated.

 

[Redbelly98]

I'm not the most rigorous mathematician (mainly because I'm a physicist), but I think (b) looks sufficient.

For (a), why not write out the expressions for O(x) and E(x) explicitly in terms of f(x)?

 

[Hurkyl]

We can now define even and odd functions such that $f(a) = O(a) + E(a)$ and $f(-a) = E(a) - O(a)$.

Why can we do that?

Sure, we would automatically know that if we knew that any funciton f with domain R can be written as the sum of an odd and even function. But isn't the exercise to actually prove that theorem?

As Redbelly98 hinted, there is an explicit formula for them, and it's pretty handy to know. (or, at least, to be able to rederive quickly)

This implies that $O_1 = (E_2 - E_1) + O_2$. However, since $O_1$ is necessarily odd and $E_2 - E_1$ is necessarily even, we have that $E_1 = E_2$ and consequently $O_1 = O_2$.

Same criticism here. This is a perfectly good argument if you know that the decomposition of a function into a sum of an odd and an even function is unique.

But the exercise is to prove the theorem, not to invoke the theorem.

 

[avec_holl]

Why can we do that?

Here's what I thought: Given any two numbers $c$ and $d$, there exist two numbers $a$ and $b$ such that $a + b = c$ and $a - b = d$. However, I guess this doesn't apply (maybe it's a consequence of this theorem?). Maybe this is incorrect in general.

As Redbelly98 hinted, there is an explicit formula for them, and it's pretty handy to know. (or, at least, to be able to rederive quickly)

I guess I'll work on this a bit longer then! Could you perhaps provide some hints to help me derive it?

Same criticism here. This is a perfectly good argument if you know that the decomposition of a function into a sum of an odd and an even function is unique.

Suppose that $O$ and $E$ are odd and even functions respectively that are not always zero. We define the function $f$ with domain $\mathbb{R}$ such that $f(x) = O(x) + E(x)$. Then $f(a) = O(a) + E(a)$ and $f(-a) = E(a) - O(a)$. Suppose now that $f$ is even, then $0 = 2O(a)$ which implies that $f$ cannot be even since $O$ is not always zero. A similar argument applies if $f$ is odd. Am I making some fatal logical error here?

Thanks for your responses!

 

[CaffeineJunky]

(a) Suppose that $f$ is some function with domain $\mathbb{R}$. This implies that for any arbitrary real number $a$, both $f(a)$ and $f(-a)$ exist. We can now define even and odd functions such that $f(a) = O(a) + E(a)$ and $f(-a) = E(a) - O(a)$. Since $f$ can be represented by the sum of even and odd functions at a single point, we can define $O$ and $E$ point-wise such that their sum is always equal to $f$.

I'm not sure if this is a reasonable argument, but if it is, how can I make it rigorous?

Here you are assuming what is to be proved--that the function is always the sum of an even and an odd function. The point of the question is to show that there always exists functions that you can define, and you assume this. (In other words, how do you know that there exists functions that cannot be written as the sum of an even and an odd function.)

Let F be a function (not the zero function) and let E be the set of terms of F that satisfy f(a) = f(-a) for all a in R. Then by definition E(x) is an even function. Show that (F - E)(x) is odd to complete the proof. (Use a proof by contradiction, the fact that the sum of two even functions is even, the sum of an even function and an odd function is neither even nor odd. Use a case by case basis: F is odd, F is even, and F is neither even nor odd. Note that the only function that is both even and odd is the zero function.)

(b) Suppose not, then $f = O_1 + E_1 = O_2 + E_2$. This implies that $O_1 = (E_2 - E_1) + O_2$. However, since $O_1$ is necessarily odd and $E_2 - E_1$ is necessarily even, we have that $E_1 = E_2$ and consequently $O_1 = O_2$. Hence, this way of writing $f$ is unique.

I'm not sure if this is a reasonable argument either and any suggestions would be appreciated.

How can you be sure from this argument that E_1 and E_2 aren't different even functions? You can't use your argument in its form because it assumes the uniqueness of the decomposition of a function into its odd and even components, which is what you are trying to prove. Instead, suppose E_1 and E_2 aren't equal to derive a contradiction that F(x) is the sum of an even function, and odd function, and another function.

 

[HallsofIvy]

You can do much better than just saying that "there exist E(x) and O(x)", you can specify exactly what they are. I will give you a hint to start: is f(x)+ f(-x) an even function? What can you do to get an odd function? Is their sum f(x)? If not, why not?

 

[avec_holl]

Alright, I'm a bit confused at the moment so I'll post my original argument with all my thoughts included . . .

(a) Suppose that $f$ is some function whose domain is the set of real numbers. This implies that both $f(a)$ and $f(-a)$ exist. Since, given any two real numbers $c$ and $d$ there exist two real numbers $a$ and $b$ such that $c = a + b$ and $d = a - b$, we define odd and even functions point-wise such that $f(a) = c = E(a) + O(a)$ and $f(-a) = d = E(a) - O(a)$. Hence, $f = O + E$.

This seems reasonable to me unless the property of numbers that I stated can only be proven using this result/theorem.


Lemma: Suppose that $O$ and $E$ are odd and even functions respectively which are not always zero. We define the function $f$ such that $f = O+E$. Now, suppose that $f$ is even, this implies that $f(a) - f(-a) = 0 = 2O(a)$. However this is a contradiction since $O$ is not always 0; hence $f$ is not even. Now suppose that $f$ is odd, this implies that $f(a) + f(-a) = 0 = 2E(a)$. However, this is also a contradiction since $E$ is not always 0; hence, $f$ is not odd. This proves that $f$ is neither odd nor even.

(b) Suppose not, then $f = E_1 + O_1 = E_2 + O_2$ which implies that $O_2 = (E_1 - E_2) + O_1$. Since $E_1 - E_2$ is necessarily even, by our first lemma we have that $E_1 = E_2$ - if not, we arrive at a contradiction - and consequently that $O_1 = O_2$.

Thanks for all the help so far! I'll post again when I have time to work through the hints. Again, thank you!

 

[CaffeineJunky]

Lemma: Suppose that $O$ and $E$ are odd and even functions respectively which are not always zero. We define the function $f$ such that $f = O+E$. Now, suppose that $f$ is even, this implies that $f(a) - f(-a) = 0 = 2O(a)$. However this is a contradiction since $O$ is not always 0; hence $f$ is not even. Now suppose that $f$ is odd, this implies that $f(a) + f(-a) = 0 = 2E(a)$. However, this is also a contradiction since $E$ is not always 0; hence, $f$ is not odd. This proves that $f$ is neither odd nor even.

This is wrong. If F is an even function, then 0 = 2O is perfectly reasonable, seeing as O is necessarily the zero function. (If O were not the zero function, then F wouldn't be even, since F = E + O would be neither odd nor even.) (But then again, I don't quite get what you were trying to prove with this lemma.)

Instead, say that since F is even, the F = E + O, where E = F and O = 0; E is even by hypothesis and O is an odd function since 0 is an odd function. (Which was to be shown.)
Now suppose that F is odd. Using a similar method, show that F can be written as the sum of an odd and even function.

The tricky part of this lemma is to show that if F is neither odd nor even then it can be written as the sum of an even and an odd function.. Do this via a proof by contraposition and contradiction: The statement you are using (rewritten) is "If F is not odd or F is not even, then F can be written as the sum of an odd function and an even function." Use the contraposition to work with an easier form to reach an easy contradiction.

Finally, note/prove that if a function is both odd and even, then the function must be the zero function, and conclude that this function can be written as the sum of an even function and an odd function. (Since you are using this previously in your argument, it may be beneficial to do this case first.)

 

[avec_holl]

This is wrong. If F is an even function, then 0 = 2O is perfectly reasonable, seeing as O is necessarily the zero function. (If O were not the zero function, then F wouldn't be even, since F = E + O would be neither odd nor even.) (But then again, I don't quite get what you were trying to prove with this lemma.)

I stated in the quoted bit that $O$ cannot be the zero function which means that $f$ cannot be even. The point of this lemma is: In (b) we have that the sum of an odd function and an even function is odd. The lemma proves that this can only be true if $E$ is the zero function - if it isn't, we arrive at a contradiction.

 

[CaffeineJunky]

Ok, I see what you did there. But you can avoid the contradiction and simply prove the theorem straightforward by using the case by case method, except for the case where F is neither even nor odd (you'd be exploiting the fact that the set of functions F that satisfy the property being contradicted ends up being empty, leading to a vacuous truth.)

EDIT: Nevermind, you proved the first part of the theorem by an alternate approach. The lemma is for the uniqueness portion.

 

[avec_holl]

Normally, I probably would use the case by case method but I'd already proved the lemma I used to derive the contradiction. So, using the lemma that I provided, do you think that (b) is a valid proof?

Also, have you read through (a)? I recognize that the point of contention is where I define the odd and even functions (and it's definitely incorrect if it's not worded properly). We know that $f(a) = a + b$ and $f(-a) = a - b$ from the analagous property of real numbers, this allows us to define the even and odd functions point-wise such that there sum is always $f$. I think that if this property can be proven independantly of this theorem, my proof should hold (or maybe I'm making another logical error somewhere).

 

[CaffeineJunky]

Your proof of A is still bad because you're still assuming that two functions E and O exist and their sum is F, where F is our arbitrary function. This defeats the purpose of showing that the functions E and O exist, since you already assume their existence. (It's easy to prove A is true when you assume A is true.)

Here's what your proof should look like.

Method 1: (This is what other people suggest. This proof is a constructive proof: you explicitly find two such functions that satisfy the properties just by using F.)
1. Take any function F : R --> A, where A is a set for which evenness and oddness can make sense.
2. Construct two functions O and E solely from F whose sum is F. (The hint here is to look at F(x) and F(-x) to construct those two functions.)

Method 2: (The one I suggest. This method is an existence proof and tells you nothing about how the functions should look.)
1. Let F be a function whose domain is R, etc.
2. Consider each case: F is even and F not odd, F is even and F is not odd, F is not even and F is odd, and F is not odd and F is not even. For each of these cases, show that F can be written as the sum of two functions.

The benefit of working through the first proof is that you can actually build two functions from F that satisfy the desired properties. The benefit of the second method is to show that you can prove that F = O + E without actually finding examples.

 

[avec_holl]

Your proof of A is still bad because you're still assuming that two functions E and O exist and their sum is F, where F is our arbitrary function. This defeats the purpose of showing that the functions E and O exist, since you already assume their existence. (It's easy to prove A is true when you assume A is true.)

I'm still struggling with this. Given any two real numbers $f(a)$ and $f(-a)$ there exists two real numbers $c$ and [/itex]d[/itex] such that $f(a) = c + d$ and $f(-a) = c - d$. The number $c$ has the property that if $E(a) = c$ then $E(-a) = c$ and the number $d$ has the property that $O(a) = d$ and $O(-a) = -d$. If this property of numbers holds without the theorem - I think it does - then I don't think I have assumed anything unreasonable. Using this property, it allows us to define point-wise two functions $O$ and $E$ that are odd and even respectively and whose sum is always $f$.

Which part of this is unreasonable? I'll also try the recommendations but first I want to understand what's wrong with my logic.

 

[CaffeineJunky]

How do you know you can always define functions O and E such that O is odd, E is even, and F = O + E? i.e. Is it possible for a function $$F \neq O + E$$ for all choices of odd functions O and even functions E? (The answer is no, but you assume that O and E will always exist without proof. This lacks rigor.)

Furthermore, you cannot assign a number c an arbitrary property: E(a) = c and E(-a) = c is not a property of c, it's a property of E. You need to find a function E such that E(a) = E(-a) for all possible choices of A--the value of E(a) is irrelevant, and a function O such that O(a) = -O(a), and then show that F(a) = E(a) + O(a) for all possible values of A. Remember, the theorem is independent of the range of the function.

 

[avec_holl]

The proof should address the first question. To address your second criticism, I'll revise the proof . . .

Suppose that $f$ is a function whose domain is the set of real numbers, then $f(a)$ and $f(-a)$ both exist. Given any two real numbers $f(a)$ and $f(-a)$ there exist two real numbers $C$ and $D$ such that $f(a) = C + D$ and $f(-a) = C - D$. We define point-wise an even function $E$ such that $E(a) = C$ and an odd function $O$ such that $O(a) = D$ (By defining these point-wise, $E(a) = E(-a)$ and $O(a) = -O(-a)$ for all real $a$). *To illuminate this last point, since given any $a$ we have that $f(a) = C + D$ and $f(-a) = C - D$ the function $O$ can be defined such that $O(a) = D$ and $O(-a) = -D$ given any real $a$. This criterion meets the definition for an odd function so we know that $O$ is odd. Similarly, we can show that the function $E$ that satisfies $E(a) = C$ and $E(-a) = C$ satisfies the definition of an even function. Hence their sum is $f(a) = O(a) + E(a)$* This implies that $f = O + E$.

Sorry if I seem stubborn but I'm really trying to understand the flaw in my logic. Thanks for your criticism!

 

[CaffeineJunky]

Your proof is nearly there, although the wording can be cleaned up.

There is a real simple solution that you are passing up that allows you to avoid using a pointwise argument and that involves simply using F(x) and F(-x), and noting a certain property about F(x) + F(-x) and F(x) - F(-x).

 

[avec_holl]

Couple thoughts that might be a bit coherent.

We know that: $f(a) = C + D$ and $f(-a) = C - D$ for some real numbers $C$ and $D$. I define the function $E$ such that is always has the property that $E(a) = C$ and $E(-a) = C$; hence $E$ is even. I define the function $O$ such that $O(a) = D$ and $O(-a) = -D$; hence $O$ is odd. All I've done with this is assigned functions that have this analagous property of the real numbers.

 

[avec_holl]

Your proof is nearly there, although the wording can be cleaned up.

There is a real simple solution that you are passing up that allows you to avoid using a pointwise argument and that involves simply using F(x) and F(-x), and noting a certain property about F(x) + F(-x) and F(x) - F(-x).

Swell! I agree that my earlier proofs weren't worded very well and that probably created some confusion. I'm sorry about that. So, as it stands, is my logic reasonable? I also realize that there are better solutions but I wanted to make sure I got everything cleared up with my logic before I proceeded (if I have/had misconceptions, I want them corrected). Thanks for all your help!

 

[CaffeineJunky]

Yes, you're logic is essentially the same as using the suggestion about F(x) + F(-x) and F(x) - F(-x).

 

[Dick]

Couple thoughts that might be a bit coherent.

We know that: $f(a) = C + D$ and $f(-a) = C - D$ for some real numbers $C$ and $D$. I define the function $E$ such that is always has the property that $E(a) = C$ and $E(-a) = C$; hence $E$ is even. I define the function $O$ such that $O(a) = D$ and $O(-a) = -D$; hence $O$ is odd. All I've done with this is assigned functions that have this analagous property of the real numbers.

That's just plain awkward. What you are doing is saying f(x)=E(x)+O(x) and f(-x)=E(x)-O(x). That's fine. But you want to show E(x) and O(x) exist. Why don't you solve those equations for E(x) and O(x) in terms of f(x) and f(-x)?

 

[Redbelly98]

That's just plain awkward. What you are doing is saying f(x)=E(x)+O(x) and f(-x)=E(x)-O(x). That's fine. But you want to show E(x) and O(x) exist. Why don't you solve those equations for E(x) and O(x) in terms of f(x) and f(-x)?

My thoughts exactly. Think of E and O as unknowns. We have a system of two equations in two unknowns. Solve for the unknowns.

And I mean these two equations, in case anybody is wondering:

$f(a) = O(a) + E(a)$ and $f(-a) = E(a) - O(a)$

Except use f(x), etc. as Dick wrote it.

 

[jgens]

My brother's not online now, but he didn't follow up on your suggestions immediately because he's been trying to clear up any misconceptions he had or has. Knowing what he means (and has meant) to say, it seems like his argument is close to the argument Redbelly98 suggests.