Is there a sense in which $\sin^2(nx)$ converges to $1/2$? Context: http://www.hairer.org/notes/Regularity.pdf, section 4.1 (pages 15-16)

Define
  $$(\Pi_x\Xi^0)(y)=1 \qquad (\Pi_x\Xi)(y)=0 \qquad (\Pi_x\Xi^2)(y)=c$$
  and
  $$(\Pi^{(n)}_x\Xi^0)(y)=1 \qquad (\Pi^{(n)}_x\Xi)(y)=\sqrt{2c}\sin(nx) \qquad (\Pi^{(n)}_x\Xi^2)(y)=2c\sin^2(nx).$$
  As a model, $\Pi^{(n)}$ converges to $\Pi$.

I don't see how this convergence is supposed to take place. Isn't the limit of $\sin(nx)$, as $n\to \infty$, undefined? 
 A: It does hold for example in the sense of periodic distributions. If $f \in L^1([0,2\pi])$ then the Riemann–Lebesgue lemma states that
$$ \lim_{n \to \infty} \int_0^{2\pi} \sin nx\, f(x)\, dx = 0\,.$$
And thus $\sin nx \to 0$ and the same holds for $\cos nx$. For $\sin^2 nx$ use simply that
$$ \sin^2 nx = \frac {1 - \cos 2nx}2\,,$$
to obtain $\sin^2 nx \to \frac 12$.
The important point, as mentioned in the comments, is that convergence comes in many different forms and shapes. As far as I understand Hairer's work on regularity structures (very little), it is about introducing new notions of convergence to make sense of objects that previously seemed or where undefined. Hence, if you want to understand his work, I would study what happened in the paper before that. Somewhere there should be an explanation or definition, how convergence is to be understood.
A: From theorem 2.10, we see that the convergence we need to check is
$$\langle \mathcal{R}^{(n)}(F_1\bigstar F_2),\psi\rangle \xrightarrow{n\to\infty} \langle \mathcal{R}(F_1\bigstar F_2),\psi\rangle.$$
I won't go over the details of how $\psi$ is defined, because they don't look relevant - the important thing is that it's compactly supported. Truth be told, we need to check the above condition with the inner product with $\psi_y^{\lambda}$ (which is slightly different to $\psi$), but it doesn't look to me like that affects the convergence, so I'll write the answer with $\psi$ instead.
We are given that
$$\mathcal{R}^{(n)}(F_1\bigstar F_2) = f_1(x)f_2(x)+\left(f_1(x)f'_2(x)+f_1'(x)f_2(x)\right)\sqrt{2c}\sin(nx)+f_1'(x)f_2'(x)2c\sin^2(nx)$$
and
$$\mathcal{R}(F_1\bigstar F_2) = f_1(x)f_2(x)+f_1'(x)f_2'(x)c.$$
From what I have understood, $\mathcal{R}^{(n)}(F_1\bigstar F_2)$ is assumed to be continuous, and hence locally integrable.
So we need to check that, as $n\to \infty$,
$$\langle \mathcal{R}^{(n)}(F_1\bigstar F_2),\psi\rangle=\int_{\mathbb{R}}f_1(x)f_2(x)\psi(x)\mathrm{d}x+\sqrt{2c}\int_{\mathbb{R}}\sin(nx)\left(f_1(x)f'_2(x)+f_1'(x)f_2(x)\right)\psi(x)\mathrm{d}x+2c\int_{\mathbb{R}}\sin^2(nx)f_1'(x)f'_2(x)\psi(x)\mathrm{d}x$$
approaches
$$\int_{\mathbb{R}}f_1(x)f_2(x)\psi(x)\mathrm{d}x+c\int_{\mathbb{R}}f_1'(x)f'_2(x)\psi(x)\mathrm{d}x.$$
The first terms are the same, so there is nothing to check there.
We need to check that the second one in $\langle \mathcal{R}^{(n)}(F_1\bigstar F_2),\psi\rangle$ goes to $0$. Calling $K$ the (compact) support of $\psi$ and using the Riemann-Lebesgue Lemma,
$$\sqrt{2c}\int_{\mathbb{R}}\sin(nx)\left(f_1(x)f'_2(x)+f_1'(x)f_2(x)\right)\psi(x)\mathrm{d}x =$$ $$ \sqrt{2c}\|\psi\|_{\infty}\int_{K}\sin(nx)\left(f_1(x)f'_2(x)+f_1'(x)f_2(x)\right)\mathrm{d}x\xrightarrow{n\to\infty}0.$$
For the third term, with $K$ as before and again using the Riemann-Lebesgue lemma,
$$2c\int_{\mathbb{R}}\sin^2(nx)f_1'(x)f'_2(x)\psi(x)\mathrm{d}x=2c\int_K\sin^2(nx)f_1'(x)f'_2(x)\psi(x)\mathrm{d}x$$
$$=2c\int_K\frac{1}{2}f_1'(x)f'_2(x)\psi(x)\mathrm{d}x-2c\int_K\frac{\cos(2nx)}{2n}f_1'(x)f'_2(x)\psi(x)\mathrm{d}x$$
$$\xrightarrow{n\to\infty}c\int_Kf_1'(x)f'_2(x)\psi(x)\mathrm{d}x = \langle \mathcal{R}(F_1\bigstar F_2),\psi\rangle.$$
