Consider the function $$f(x) = \frac{1}{\sqrt{x}}\mathbf{1}_{[0,1]}(x),$$ where $\mathbf{1}_{[0,1]}$ is the indicator function that is equal to $1$ on the set $[0,1]$ and $0$ otherwise. We want to show that $\hat{f}\not\in L^{1}(\mathbb{R})+L^{2}(\mathbb{R}).$
My Attempt: First we compute the Fourier transform $$\hat{f}(\zeta) = \int_{\mathbb{R}} e^{-i\zeta x} f(x)dx = \int_{[0,1]} \frac{e^{-i\zeta x}}{\sqrt{x}}dx = 2\int_{[0,1]} e^{-i\zeta t^2}dt.$$ Now if by contradiction suppose there exists $f_1\in L^1(\mathbb{R})$ and $f_2\in L^2(\mathbb{R})$ such that, $$\hat{f} = f_1 + f_2.$$ Then I am not sure how to use this to derive a contradiction. We know that $||f_1||_{L^1}<\infty$ and $||f_2||_{L_2}<\infty$ but then how do we use this information to solve the problem? Any hints would be much appreciated.
Assume that $\hat{f} \in L^1(\mathbb{R})+L^2(\mathbb{R})$, then we can write $\hat{f}=g+h$ with $g\in L^1(\mathbb{R})$ and $h\in L^2(\mathbb{R})$. This gives $f=\check{g}+ \check{h} \in L^\infty(\mathbb{R})+L^2(\mathbb{R})$. As $f$ has support in $[0,1]$, we can write $$f= 1_{[0,1]} \check{g} + 1_{[0,1]} \check{h}.$$ This implies $f\in L^2(\mathbb{R})$ (as $L^\infty([0,1])\subset L^2(\mathbb{R})$), which gives the desired contradiction.