I am wondering if it is allowed to interchange the limit and integral sign in the following expression:
$$\lim\limits_{N\rightarrow\infty}\int\limits_{-\infty}^{\infty} \frac{1}{2\pi}\sum_{k=-N}^{N} b_ke^{ikx} \varphi(x) dx.$$
Here, $\varphi$ denotes a Schwartz function in $\mathbb{R}$ and $\frac{1}{2\pi}\sum\limits_{k=-\infty}^{\infty} b_ke^{ikx}$ denotes the Fourier transform of a periodic continuous function $f:\mathbb{R}\rightarrow\mathbb{R}$ with period $2\pi.$ By definition $b_k:=\int_0^{2\pi}f(y)e^{iky}dy.$
Unfortunately, the usual theorems like dominated or monotone convergence can't be applied here. But my feeling is that the limit above should equal to
$$\int\limits_{-\infty}^{\infty} \frac{1}{2\pi}\sum_{k=-\infty}^{\infty} b_ke^{ikx} \varphi(x) dx$$
How can this be justified? Thank you for your support.
Best Answer
The integral and the limit can indeed be interchanged. To see this, first note that if $g \in L^2_{\mathrm{loc}}(\Bbb{R})$ is $2\pi$-periodic, then \begin{align*} \int_{\Bbb{R}} g(x) \, \varphi(x) \, d x & = \sum_{k \in \Bbb{Z}} \int_{2\pi k}^{2\pi k+2\pi} g(x) \, \varphi(x) \, d x \\ & = \sum_{k \in \Bbb{Z}} \int_{0}^{2\pi} g(x) \, \varphi(x + 2\pi k) \, d x \\ & = \int_0^{2\pi} g(x) \, P \varphi (x) \, d x, \end{align*} where $P \varphi(x) = \sum_{k \in \Bbb{Z}} \varphi(x + 2\pi k)$ is bounded (since $\varphi$ is a Schwartz function) and $2\pi$-periodic. The interchange of the series and the integral above can be easily justified using the dominated convergence theorem.
Now, define $f_N := \frac{1}{2\pi} \sum_{k=-N}^N b_k e^{i k x}$, and note that $f_N \to f$ in $L^2([0,2\pi])$. Since $P \varphi \in L^\infty([0,2\pi]) \subset L^2 ([0,2\pi])$, we see \begin{align*} \int_{\Bbb{R}} f_N (x) \, \varphi(x) \, d x & = \int_0^{2\pi} f_N (x) \, P \varphi (x) \, d x \\ & \xrightarrow[N\to\infty]{} \int_0^{2\pi} f(x) \, P \varphi (x) \, d x \\ & = \int_{\Bbb{R}} f(x) \, \varphi(x) \, d x , \end{align*}
which is what you wanted to show.