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Consider a continuous $L^1$ function $f$ : $\mathbb{R}$ $\rightarrow$ $\mathbb{R}$ such that $supp$ $\widehat{f}$ $\subset$ $[-1,1]$ and $f(n)$ $\geq$ $0$ if $n$ $\in$ $\mathbb{Z}$. The problem is to prove that the integral of $f$ over $\mathbb{R}$ is non-negative.

What I've done: The Poisson summation formula holds a.e., i.e., $$ F(x) = \sum_ {\mathbb{Z}} f(x+n)= \sum_{\mathbb{Z}}\widehat{f}(n)e^{2\pi inx} = \widehat{f}(0) = \int_{\mathbb{R}}f(y)dy $$ almost everywhere. We can find a sequence $x_{n}$ $\rightarrow$ $0$ such that this formula holds, but I can't garantee that $F$ is continuous. For this to be true, the series $\sum_ {\mathbb{Z}} f(x+n)$ would have to be uniformly convergent so $F$ would be an uniform limit of continuous functions. This holding, I just have to take the limit when $x$ $\rightarrow$ $0$ and use the hypothesis that $f$ is non-negative on integers. What am I missing?

Br09
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1 Answers1

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Hopefully, third time's the charm.

Goal Let $f\in L^{1}(\mathbb{R})$ be continuous function, such that $\text{supp}(\hat{f})\subset [-1,1]$ and $f(n)\geq 0$ for all $n\in\mathbb{Z}$. Then $$\int_{\mathbb{R}}f(y)dy\geq 0$$

Set $g(x):=\hat{f}(-x)=f^{\vee}(x)$. Clearly, $\text{supp}(g)\subset [-1,1]$, and since $g$ is continuous, $x^{n}g(x)\in L^{1}(\mathbb{R})$ for any $n\in\mathbb{N}$. By Fourier inversion and the hypothesis that $f$ is continuous, we see that $\hat{g}(\xi)=f(\xi)$ everywhere. Define a function $G:\mathbb{T}\rightarrow\mathbb{C}$ by $$G(x)=\sum_{n\in\mathbb{Z}}g(x+n)=\sum_{n\in\mathbb{Z}}\hat{f}(-x-n)$$

I claim that the series defining $G$ converges uniformly and absolutely. Indeed, for any $x\in[0,1]$ and $n>2$, we have that $\left|x+n\right|>1$, whence $\hat{f}(-x-n)=0$. Since $\hat{f}$ is continuous, $G$ is the finite sum of continuous functions and therefore continuous.

I claim that $\hat{G}(m)=\hat{g}(m)=f(m)$ for all $m\in\mathbb{Z}$. Indeed, the second equality was established above, so it remains for us to prove the first. $$\int_{0}^{1}G(x)e^{-2\pi i mx}dx=\sum_{n\in\mathbb{Z}}\int_{0}^{1}g(x+n)e^{-2\pi i m x}dx=\sum_{n\in\mathbb{Z}}\int_{n}^{n+1}g(x)e^{-2\pi i m(x-n)}dx=\int_{\mathbb{R}}g(x)e^{-2\pi i mx}dx$$

To complete the proof we need, we'll need a classical theorem of Fejer.

Theorem Denote the $N^{th}$ symmetric partial sum of the Fourier series of a continuous function $h:\mathbb{T}\rightarrow\mathbb{C}$ by $$S_{N}h(x)=\sum_{\left|m\right|\leq N}\hat{h}(m)e^{2\pi i m x}, x\in\mathbb{T}$$ and define the $N^{th}$ Cesaro mean of the partial sums of the Fourier series of $h$ by $$\sigma_{N}h(x)=\dfrac{1}{N}\sum_{j=0}^{N-1}S_{j}h(x), \ x\in\mathbb{T}$$

Then $\sigma_{N}h$ converges to $h$ uniformly on $\mathbb{T}$.

Using Fejer's theorem with $h=G$, we have in particular that $\sigma_{N}G(0)\rightarrow G(0)$. Since $S_{j}G(0)=\sum_{\left|m\right|\leq j}f(-m)=\sum_{\left|m\right|\leq j}f(m)$, it follows that $\sigma_{N}G(0)\geq 0$, whence

$$\int_{\mathbb{R}}f(y)dy=\hat{f}(0)=G(0)=\lim_{N\rightarrow\infty}\sigma_{N}G(0)\geq 0$$

Matt Rosenzweig
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