This was a typo. Sub

and

, in order to change the limits of integration from

to

in accordance with the R-L lemma. Thanks for catching that blooper, Sampras.

Yes it does. Completing the RHS of the substitution,

Regardless of what f is, it needs to sit on the origin. Here is a proof: If

for all

, then h must be an even function, that is, symmetric about the y-axis. So, we can think of our integral in question as "tending towards" an even function as n grows large. Obviously, there does not have to be (probably won't be) an n large enough that

but

does need to be zero.

Consider this counterexample:

, so

Theorem:

Proof:

=

=

*I'm not sure how rigorous this is, but because the

is not tending to an even function as n gets large, the area bounded under it on either side of the y-axis will always be inequal for some choice of L.