Dominated convergence to characteristic function

Question

Let $\phi_m(x):=\chi_{[0,1]} * \chi_{[0,1]} *...* \chi_{[0,1]}$ be the m -times convolution (so $m+1$ characteristic functions are involved).

Then the Fourier transform of this function is given by

$$\hat{\phi}_m(x):=\left( \frac{e^{-ix}-1}{-ix} \right)^{m+1}.$$

Now, I want to show that for

$$ \hat{f}_m(x):= \frac{\hat{\phi}_m(x)}{\sum_{l \in \mathbb{Z}} |\hat{\phi}_m(x+2\pi l)|^2}$$

we have $|| \hat{f}_m- \chi_{[-\pi,\pi]}||_{L^2} \rightarrow 0.$

I already showed that $| \hat{f}_m| \rightarrow \chi_{[-\pi,\pi]}$ pointwise and that there exists $|g| \in L^2$ such that $|\hat{f}_m| \le |g|.$

So the proposition would immdiately follow, if I would know that the imaginary part of this $\hat{f}_m$ would tend to zero for $m \rightarrow \infty$ (and I need this result only for $|x| \le \pi.$ Unfortunately, I don't see how this can be done. I tried expanding the product, but this was just messy.

Does anybody have an idea? Maybe there is more abstract reason, why the imaginary part has to vanish in the limit?

If you have any questions, please let me know in the comment section. In particular, any other possible proof of this statement is also highly appreciated.

Answer 1

I think the sequence does not converge in $L^2.$

We have $$\vert \hat f_{4m-1}(x)\vert= e^{i2xm}\hat f_{4m-1}(x)= {1\over{(2x\sin(x/2))^{4m}\sum_{l\in Z} {1\over{(x+2\pi l)^{8m}}}}}$$ Let us suppose $ \lim \hat f_m = f ,\ \lim \vert \hat f_m\vert = \vert f \vert$ in $L^2$.

If we define $g_m(x)=e^{2imx}f(x),$ we have

$$\lim\Vert g_m-\vert \hat f_{4m-1}\vert\Vert_2 =\lim\Vert g_m- \vert f\vert\Vert_2 =0 \ \Rightarrow \ \ \lim\Vert g_{m}- g_{2m}\Vert_2 =0 .$$ If $\vert f\vert >\epsilon $ on some bounded interval $I$ we deduce $$\lim_m\int_I \sin^2 (m x) \vert f(x)\vert^2dx=\lim_m\int_I \sin^2 (m x) dx=0$$ which is a contradiction.