Compactly supported Fourier transform for "patching function"

Question

Let $g:\mathbb{R}\backslash\{0\}\to\mathbb{R}$ be the function $g(x)=\frac{1}{x^2}$ and let $I=[-c,c]$, for some $c>0$. I would like to know if there exists a function $\varphi:\mathbb{R}\to\mathbb{R}$ satisfying the following properties:

1. $\varphi$ and $g$ coincide in $\mathbb{R}\backslash[-c,c]$.
2. The Fourier transform of $\varphi$ is compactly supported.

Property 2 is clearly the hardest one to achieve: not only does it entail that $\varphi$ should be smooth (so we can't just glue the function $\frac{1}{c^2}\chi_{[-c,c]}$ to $g$), but requiring the Fourier transform to have compact support forces $\varphi$ to be (real) analytic. Apparently, 2) is equivalent to $\varphi$ being the restriction to $\mathbb{R}$ of an entire (complex) function with some decay behaviour on the imaginary part, but I can't even go past finding a real analitic function satisfying the properties, to start with.

To tackle this problem, I tried using bump functions (big mistake, in hindsight): let $$\varphi(x)= \begin{cases} g(x)(1-e^{1-1/(1-(x/c)^2)}), & x\in [-c,c],\\ g(x),& x\notin[-c,c]. \end{cases} $$ Now, although this is a smooth function satisfying 1), the derivatives of $\varphi$ and $g$ coincide at $\pm 1$. Therefore, the Taylor expansions for $\varphi$ centered at $\pm 1$ do not converge to $\varphi$, but to $g$, rather. Thus, $\varphi$ is not analytic, which prevents 2).

A proof of existence (or non-existence) of such function would be enough, but I have been trying to cook such $\varphi$ up nonetheless. Any help is appreciated! \begin{equation} \end{equation}

Answer 1

Suppose that $\varphi$ that meets those conditions exists. Then $\varphi$ can be extended as an entire function of the complex variable (holomorphic/analytic on the whole complex plane). That's because you can write $$\varphi(x)=\int_{-a}^a\hat \varphi(\xi)e^{2i\pi \xi x }d\xi$$ thus you can build the extension $$\varphi(z)=\int_{-a}^a\hat \varphi(\xi)e^{2i\pi \xi z }d\xi=\sum_{n\geq 0}\frac{(2i\pi z)^n}{n!}\int_{-a}^a\hat\varphi(\xi)\xi^n d\xi$$

Now, for any $x\in \mathbb R$, define $h(x)=x^2\varphi(x)$. Then $h$ inherit the same property as $\varphi$: You can extend it as an entire function of the complex variable. But by definition, $h(x)=1$ for all $|x|> c$. This implies that $h(x)=1$ for all $x\in\mathbb R$.

So it is not possible.