How to prove that inverse Fourier transform of "1" is delta funstion?

In the following $\langle f, \cdot \rangle$ denotes the linear functional on Schwartz space induced by $f$ and $f^\lor$ stands for the inverse Fourier transform of $f$. By definition, for any Schwartz function $\varphi$ \begin{align*} \langle 1^\lor, \varphi \rangle=\langle 1, \varphi^\lor \rangle=&\int_\mathbb{R} \left(\int_\mathbb{R} e^{2\pi ixy}\varphi(y) dy\right)dx =\lim_{M\to\infty}\int_{-M}^M \left(\int_\mathbb{R} e^{2\pi ixy}\varphi(y) dy\right)dx. \end{align*} By Fubini's theorem we have \begin{align*} \int_{-M}^M \left(\int_\mathbb{R} e^{2\pi ixy}\varphi(y) dy\right)dx=&\int_\mathbb{R} \varphi(y)\left(\int_{-M}^M e^{2\pi ixy}dx\right) dy =\pi^{-1}\int_\mathbb{R} \varphi\left(y\right)\frac{\sin (2\pi My)}{y} dy. \end{align*} Since $\varphi$ is differentiable at $y=0$, we have $|\varphi(y)-\varphi(0)|\le C|y|$ for some constant $C$. Thus $\left(\varphi(y)-\varphi(0)\right)/y\in L^1_{loc}(\mathbb{R}).$ Then by Riemann-Lebesgue Lemma we have \begin{align*} \lim_{M\to\infty}\int_{-1}^1 \left(\varphi(y)-\varphi(0)\right)\frac{\sin (2\pi My)}{y} dy=0, \end{align*} which means \begin{align*} \lim_{M\to\infty}\int_{-1}^1 \varphi\left(y\right)\frac{\sin (2\pi My)}{y} dy=&\varphi(0)\lim_{M\to\infty}\int_{-1}^1\frac{\sin (2\pi My)}{y} dy \\ =&\varphi(0) \int_{-\infty}^\infty \frac{\sin y}{y}dy=\pi\varphi(0). \end{align*} Note that $\varphi\left(y\right)/y$ is integrable on $\mathbb{R}\setminus [-1,1]$. Thus by Riemann-Lebesgue Lemma we have \begin{align*} \lim_{M\to\infty}\int_{1}^\infty \varphi\left(y\right)\frac{\sin (2\pi My)}{y} dy=\lim_{M\to\infty}\int_{-\infty}^{-1} \varphi\left(y\right)\frac{\sin (2\pi My)}{y} dy=0. \end{align*} To sum up the above argument we have for all Schwartz functions $\varphi$, \begin{align*} \langle 1^\lor, \varphi \rangle=\pi^{-1}\lim_{M\to\infty}\int_\mathbb{R} \varphi\left(y\right)\frac{\sin (2\pi My)}{y} dy=\varphi(0)=\langle \delta, \varphi \rangle. \end{align*} Therefore $1^\lor=\delta$.