Suppose $\lim_n e^{itb_n}$ exists for all $|t|\le \delta$, show that $\limsup |b_n| < \infty$

Question

I would like some help with the following:

Let $(b_n)_{n \ge 1}$ be a sequence of numbers such that $\lim_n e^{itb_n}$ exists for all $|t|\le \delta$, $\delta >0$. Show that $\limsup |b_n| < \infty$.

Edit: as per Mindlack's hint, below is an attempt to a solution.

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For each $n$ define $$ f_n(t) = e^{itb_n}, \quad |t| < \delta. $$ Then, by assumption, $(f_n(t))_n$ converges for every $|t|\le \delta$. Obviously each $f_n$ is square integrable since $$ \int_{-\delta}^{\delta} \left | f_n \right |^2 \, d \mu = \int_{-\delta}^{\delta} 1 \, d \mu = 2\delta. $$ As pointwise convergence implies convergence in $L^2$, and $L^2([-t,t], \mu)$ is a complete metric space, the limit is also square integrable.

To calculate the Fourier transform of this limit, one may use the Dominated convergence theorem. \begin{align*} \int_{-\delta}^{\delta} e^{-2 \pi it \xi}\lim_n f_n(t) \, d \mu(t) &= \lim_n\int_{-\delta}^{\delta} e^{-2 \pi it \xi}f_n(t) \, d \mu(t) \\ &= \lim_n\int_{-\delta}^{\delta} e^{-2 \pi it \xi}e^{itb_n} \, d \mu(t) \\ &= \lim_n\int_{-\delta}^{\delta} e^{it(b_n -2 \pi \xi)}\, d \mu(t)\\ &= \lim_n \left (\frac{e^{i\delta(b_n -2 \pi \xi)}}{i(b_n -2 \pi \xi)} - \frac{e^{-i\delta(b_n -2 \pi \xi)}}{i(b_n -2 \pi \xi)} \right ) \\ &= \lim_n \frac{2i \sin(\delta(b_n -2 \pi \xi))}{i(b_n -2 \pi \xi)} \\ &= \lim_n \frac{2 \sin(\delta(b_n -2 \pi \xi))}{b_n -2 \pi \xi}. \end{align*}

I'm not sure what conclusion I could draw from this?

Thanks in advance!

Answer 1

For any $f\in L_1$, denote by $\hat{f}$ its Fourier transform.

In your notation, let $g(t)=\lim_n f_n(t)$, where $f_n(t)=e^{ib_n t}\mathbb{1}_{[-\delta,\delta]}(t)$. Each $f_n\in L_1(\mathbb{R})$ and $\|f_n\|_1=2\delta>0$. Then by dominated convergence $\|g\|_1=2\delta>0$. Suppose there is a subsequence $|b_{n_k}|\xrightarrow{k\rightarrow\infty}\infty$. Then (dominated convergence) $$\hat{g}(\xi)=\lim_k\hat{f_{n_k}}(t)=2\lim_k\frac{\sin\big((b_{n_k}-2\pi\xi)\delta\big)}{b_{n_k}-2\pi\xi}=0$$ This means that $g(t)\equiv0$, contradicting the fact that $\|g\|_1>0$.

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One can also extend $f_n$ as $2\delta$-periodic functions and use Fourier series instead, The ideas is similar. The $m$-th Fourier coefficient of $g$ is the limit of the $m$-th Fourier coefficient of $f_n$. Arguing as before, you get that $g\equiv0$ and get a contradiction to the positivity of $\|g\|_{L_2(\mathbb{T}^1) }$.