I am trying to prove that if $\phi \in L_{1,loc} (\mathbb{R^d}), \phi(x)\geqslant 0$, and $\int_{\mathbb{R^d}}\phi(x)dx=\infty$, then Fourier transform $\hat\phi$ is unlimited.
My try
By contradiction. Suppose that $\hat\phi$ is an limited function. Than $\lvert {\int_{\mathbb{R^d}}\hat\phi(\xi)u(\xi)d\xi}\rvert\leqslant c{\lVert u\rVert}_{L_1}$
Because $\int_{\lvert x \rvert<R}\phi(x)dx\to\infty$, then $\int_{\mathbb R^d}\phi(x)u(\frac xR)dx\to\infty$, where $u(x)= \begin{cases} 1 &\text{if } \lvert x \rvert<1\\ 0 &\text{if } \lvert x \rvert>2 \end{cases}$
If we express the first expression through the second one, we will come to a contradiction.
I tried to use the theorems about changing the order of integration to express the first integral in terms of $\int_{\mathbb R^d}\phi(x)\hat u(x)dx$ but these theorems are not applicable in this problem. Could you suggest a way to solve this problem?