Limit - Distributions

[Camillo]

Determine the limit : $ lim_(k to oo ) sin^2kx $ in $ D'(RR) $

[Kroldar]

To each term of the sequence of functions $sin^2(kt)$ we can associate a distribution, in order to have a sequence $(T_k)$ of distributions.
We want to find the limit $T$ of this sequence.

First of all, we can say that $(T_k) to T$ in $ccD'$ if $langle T,phi rangle = lim_k langle T_k,phi rangle$, $AA phi in ccD$.

It results
$langle T_k,phi rangle = int_(-oo)^(+oo) sin^2(kt) phi(t) dt$, $AA phi in ccD$.
Let's observe that what we wrote above is not an abuse of notation, because $sin^2(kt) in L_(loc)^1$, so $T_k$ is a sequence of regular distributions.

We can also write $sin^2(kt)$ in a different way, in fact $sin^2(kt) = (1-cos(2kt))/2$ and we have
$langle T_k,phi rangle = int_(-oo)^(+oo) sin^2(kt) phi(t) dt = int_(-oo)^(+oo) (1-cos(2kt))/2 phi(t) dt = int_(-oo)^(+oo) 1/2 phi(t) dt - int_(-oo)^(+oo) cos(2kt)/2 phi(t) dt$, $AA phi in ccD$.

We obtain
$lim_k langle T_k,phi rangle = lim_k int_(-oo)^(+oo) 1/2 phi(t) dt - lim_k int_(-oo)^(+oo) cos(2kt)/2 phi(t) dt$, $AA phi in ccD$.

The first integral does not depend on $k$.

Let's calculate by parts the second integral
$lim_k int_(-oo)^(+oo) cos(2kt)/2 phi(t) dt = lim_k ([(sin(2kt))/(4k) phi(t)]_(-oo)^(+oo) - 1/(2k) int_(-oo)^(+oo) sin(2kt) phi'(t) dt)$.
That's simple to observe that
$[(sin(2kt))/(4k) phi(t)]_(-oo)^(+oo) = 0$ because $phi$ has compact support
and also
$|int_(-oo)^(+oo) sin(2kt) phi'(t) dt| <= int_(-oo)^(+oo) |phi'(t)| dt = M < +oo$
so it results
$lim_k (M)/(4k) = 0$
and we find that the integral is $0$ when $k to +oo$.

Finally
$lim_k langle T_k,phi rangle = int_(-oo)^(+oo) 1/2 phi(t) dt = langle 1/2,phi rangle$, $AA phi in ccD$.

The limit of the sequence is still a regular distribution.

[Camillo]

That's right , it would be appreciated if you give the explanation of how you get that

[Kroldar]

I think that $sin^2(kt) to 1/2$ in $ccD'$.