Problem about Fourier transform being integrable

Question

I am currently reading a paper and the author makes the following claim: If $f \in L^1(\mathbb{R})$ is a continuous, even, and nonnegative function such that $\hat{f}(\alpha) \leq 0$ for $|\alpha| \geq 1$, then $\hat{f} \in L^1(\mathbb{R})$. He claims that this can be shown by approximation of the identity. I am not very comfortable with approximation of the identity. I would be extremely grateful if a proof could be outlined.

Answer 1

It suffices to show that the integral $$\int\limits_{|t|\ge 1} [-\widehat{f}(t)]\,dt$$ is convergent. To this end we will show that the integrals $$\int\limits_{|t|\ge 1} [-\widehat{f}(t)]e^{-a t^2}\,dt$$ are bounded with respect to $a>0.$ As the function $\widehat{f}$ is continuous the latter is equivalent to the boundedness of the integrals $$\int\limits_{-\infty}^\infty \widehat{f}(t)e^{-a t^2}\,dt$$

We have $$\int\limits_{-\infty}^\infty \widehat{f}(t)e^{-at^2}\,dt =\int\limits_{-\infty}^\infty\left ( \int\limits_{-\infty}^\infty f(x)e^{-2\pi it x}e^{-at^2}\,dx\right )\,dt\\ \int\limits_{-\infty}^\infty f(x)\left (\int\limits_{-\infty}^\infty e^{-at^2}e^{-2\pi itx}\,dt\right )\,dx\ = \sqrt{\pi\over a}\int\limits_{-\infty}^\infty f(x)e^{-\pi^2x^2/a}\,dx $$ The change of integration was justified as the function $f(x)e^{-2\pi itx} e^{-at^2}$ is absolutely integrable over $\mathbb{R}^2.$ The last integral can be split into $$ \sqrt{\pi\over a}\int\limits_{|x|\le 1} f(x)e^{-\pi^2x^2/a}\,dx+\sqrt{\pi\over a}\int\limits_{|x|\ge 1} f(x)e^{-\pi^2x^2/a}\,dx\\ \le \max_{|x|\le 1}f(x) \ \sqrt{\pi\over a}\int\limits_{|x|\le 1} e^{-\pi^2x^2/a}\,dx+\sqrt{\pi\over a}e^{-\pi^2/a}\int\limits_{|x|\ge 1}f(x)\,dx\\ \le \max_{|x|\le 1}f(x)+\sup_{y>0}ye^{-y^2}\,\|f\|_1\\ = \max_{|x|\le 1}f(x)+{1\over \sqrt{2}}e^{-1/2}\|f\|_1$$

Remark The function $f$ does not to be even. Morevoer its nonnegativity is not essential as in the last part we can estimate the integrals using $|f(x)|.$