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I woke up this morning and had this question in mind. Just curious if such function can exist.

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In case anyone has forgotten what “even” and “odd” functions are, $f$ is even if $f(x) = f(-x)$ and odd if $-f(x) = f(-x)$. See also Wikipedia on even and odd functions. – Rory O'Kane Jun 17 '12 at 18:54
You might find it interesting that I often used to ask this as an extra credit question on precalculus tests when even/odd function properties were covered, typically worth an extra 3 points on a 100 point scale (so a score of 103/100 was possible). I'd usually get about 2 to 5 students getting the extra points (out of a total of maybe 25-35 students) in a U.S. college precalculus class, and about half the class getting the extra points in U.S. honors level high school classes I used to teach. – Dave L. Renfro Jun 18 '12 at 15:56
up vote 45 down vote accepted

Others have mentioned that $f(x)=0$ is an example. In fact, we can prove that it is the only example of a function from $\mathbb{R}\to \mathbb{R}$ (i.e a function which takes in real values and outputs real values) that is both odd and even. Suppose $f(x)$ is any function which is both odd and even. Then $f(-x) = -f(x)$ by odd-ness, and $f(-x)=f(x)$ by even-ness. Thus $-f(x) = f(x)$, so $f(x)=0.$

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Of course, one could argue that restrictions of the constant $0$ function to different domains symmetric about the origin are different functions, set-theoretically speaking. – Cameron Buie Jun 17 '12 at 15:30
@CameronBuie That is true, I will make my answer more precise to indicate this. Thank you. – Ragib Zaman Jun 17 '12 at 15:31
Funny, I never thought of f(x) = 0 as a possibility. Thanks for the answers everyone! – bodacydo Jun 17 '12 at 21:06

If $K$ is a field of characteristic 2, every function $K\to K$ is both even and odd.

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i'm sorry, wouldn't that be "unequal to 2"? – akkkk Jun 17 '12 at 15:31
@Auke: No. I won't spoil the joke by spelling it out, sorry. – Harald Hanche-Olsen Jun 17 '12 at 15:40
Actually, you don't even need a field, any ring of characteristic 2 will do. – Ilmari Karonen Jun 17 '12 at 15:42
This is a wonderful answer! – Edward Hughes Jun 17 '12 at 23:52
@Agos: Have a look at this wikipedia page, and try to work out the consequences of the characteristic being 2. – Harald Hanche-Olsen Jun 18 '12 at 11:33

Yes. The constant function $f(x) = 0$ satisfies both conditions.

Even: $$ f(-x) = 0 = f(x) $$

Odd: $$ f(-x) = 0 = -f(x) $$

Furthermore, it's the only real function that satisfies both conditions:

$$ f(-x) = f(x) = -f(x) \Rightarrow 2f(x) = 0 \Rightarrow f(x) = 0 $$

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Hint $\rm\ f\:$ is even and odd $\rm\iff f(x) = f(-x) = -f(x)\:\Rightarrow\: 2\,f(x) = 0.\:$ This is true if $\rm\:f = 0,\:$ but may also have other solutions, e.g. $\rm\:f = n\:$ in $\rm\:\mathbb Z/2n =\:$ integers mod $\rm 2n,$ where $\rm\: -n \equiv n.$

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+1, but note that your last $\iff$ applies (in the backwards, i.e. 'if' direction) only to $f(x) = -f(x)$, and not to the part where $f(-x)$ equals both of them. – ShreevatsaR Jun 17 '12 at 18:04
Yes, I meant to write $\:\Rightarrow\: $ but it was lost in editing. Now fixed. Thanks. – Bill Dubuque Jun 17 '12 at 18:19

Suppose $f$ odd an even. Let $x \in D$ ( D is set definition of $f$) then you have : $ f(x)=f(-x)=-f(x)$. What can you conclude about $f$ ?

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As other people have mentioned already, the real function $f(x)$ which maps every real number to zero (i.e.$f(x) = 0 \space \forall x \in \mathbb{R}$) is both even and odd because $$f(x) - f(-x) = 0 \space \space , f(x)+f(-x) = 0\space \forall x \in \mathbb{R} .$$ Also it is the only function defined over $\mathbb{R}$ to possess this property.

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Let $R$ be a Boolean ring and $X$ be an arbitrary set. Then every function $f:R\rightarrow X$ is both even and odd.

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Say what........? – bodacydo Sep 9 '15 at 7:06
@bodacydo for example: any polynomial $P(x)\in R[x]$. – user48941 Sep 9 '15 at 7:31
Thanks that is useful to know! – bodacydo Sep 9 '15 at 20:12

I post this as an addendum to the awesome answers already present for this Q.

Most trivial example:

For $f(x)=0$ we have:

$f(x)= 0 =-f(-x)$ Hence, Odd

$f(x)= 0 =f(-x)$ Hence, Even

I was searching for an answer to my question.

$f(x) = \sin(x) + \sin(\pi + x)$

Here, the function is also even and odd at the same time (another example or representation of the same). This is due to the value resulting in zero over the entire domain. So, it can be concluded that all the functions that have their $Range=\{0\}$ should be both even and odd at the same time despite their notation is of a constant function($f(x)=0$) or not.

Another example,

$f:\{1,-1\}\to \mathbb{R}$

$f(x) = x^{2}-1$

$f(x) = 0$ for all values in the domain. So, it is both even and odd at the same time because while deciding even odd functions the domain to which the function is restricted by definition must be considered.

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