Is the Intersection of Even and Odd Function Subspaces Only the Zero Function?

[vdgreat]

Let V be the vector space of all functions from R to R, equipped with the usual
operations of function addition and scalar multiplication. Let E be the subset of even
functions, so E = {f $$\epsilon$$ V |f(x) = f(−x), $$\forall$$x $$\epsilon$$ R} , and let O be the subset of odd
functions, so that O = {f $$\epsilon$$ V |f(x) = −f(−x), $$\forall$$x $$\epsilon$$ R} . Prove that:
(a) E and O are subspaces of V .
(b) E $$\cap$$ O = {0}.
(c) E + O = V .

 

[grmnsplx]

This is a good problem.

I don't think this should be too difficult for you. I think all you need is a few tips to help you get it. How about you show us what you have done so far?

Basically, all you need to do is apply your definitions. and use one of your proof techniques. For example, in "b" try using proof by contradiction. So assume that th intersection of E and O is not empty, and see where this leads you.

Give the problem a try or tell us your ideas at least and then someone will give you some good comments and feedback. :)

 

[vdgreat]

i did a)
trying b and see where it leads

thanks

 

[vdgreat]

for b)

i said
E$$\cap$$O= {f$$\epsilon$$v | f(x)= f(-x) and f(x)=-f(-x), $$\forall$$x$$\epsilon$$R}
E$$\cap$$O= {f$$\epsilon$$v | f(x)= -f(x), $$\forall$$x$$\epsilon$$R}
E$$\cap$$O= {f$$\epsilon$$v | f(x)= 0, $$\forall$$x$$\epsilon$$R}
E$$\cap$$O= {0} (f(x) =0 is the zero function)

what do you think?

 

[vdgreat]

do we need contradiction?
and how about c) any idea??

 

[HallsofIvy]

Be careful with c) E+ O is the space spanned by the vectors in E and in O. In other words, functions in it can be written as a linear combination of even and odd functions.

You might want to think about this: Given any function f, we can define
a) $f_E(x)= (f(x)+ f(-x))/2$, the even "part" of f
b) $f_O(x)= (f(x)- f(-x))/2$, the odd "part" of f

Can you see that $f_E$ is an even function and that $f_O$ is an odd function? And that $f(x)= f_E(x)+ f_O(x)$? What does that tell you about part c?

 

[grmnsplx]

for b)

i said
E$$\cap$$O= {f$$\epsilon$$v | f(x)= f(-x) and f(x)=-f(-x), $$\forall$$x$$\epsilon$$R}
E$$\cap$$O= {f$$\epsilon$$v | f(x)= -f(x), $$\forall$$x$$\epsilon$$R}
E$$\cap$$O= {f$$\epsilon$$v | f(x)= 0, $$\forall$$x$$\epsilon$$R}
E$$\cap$$O= {0} (f(x) =0 is the zero function)

what do you think?

Not bad. But it depends on how anal your prof is.
I think you might want to use contradiction just to be safe here. (Someone else please tell me if they disagree).

We have that the zero function is in O and E by (a). Now assume that there is another function in the intersection which is NOT equal to the zero function. Then show that is is incorrect by your argument above.

that seems the safest way.