scaling property and Fourier transform of Dirac distribution

Question

Let $$\delta_x:\mathcal S(\mathbb R^d)\to\mathbb R\;,\;\;\;\varphi\mapsto\varphi(x)$$ for $x\in\mathbb R^d$. If $x\in\mathbb R^d$ and $t>0$, the Fourier transform of $\delta_{tx}$ is given by $$\langle\varphi,\hat\delta_{tx}\rangle=\langle\hat\varphi,\delta_{tx}\rangle=\int e^{-{\rm i}2\pi\langle y,\:tx\rangle}\varphi(y)\:{\rm d}y=\int\frac1{t^d}e^{-{\rm i}2\pi\langle y,\:x\rangle}\varphi\left(\frac yt\right)\:{\rm d}y\tag1$$ for all $\varphi\in\mathcal S(\mathbb R^d)$. However, from the identity for the Dirac comb (entry 314) it seems like we should have $$\hat\delta_{tx}=\frac1{t^d}\delta_{\frac xt}\tag2.$$ So, what am I missing?

Answer 1

$$\langle\varphi,\hat\delta_{tx}\rangle=\langle\hat\varphi,\delta_{tx}\rangle=\hat{\varphi}(tx)=\int_{\Bbb{R}^d} e^{-{\rm i}2\pi\langle y,\:tx\rangle}\varphi(y)\:{\rm d}y=\langle \varphi,e^{-{\rm i}2\pi\langle y,\:tx\rangle}\rangle$$

$\sum_{x\in \Bbb{Z}^d}\hat\delta_{tx}=\sum_{x\in \Bbb{Z}^d}\frac1{t^d}\delta_{\frac xt}$ doesn't mean that $\hat\delta_{tx}=\frac1{t^d}\delta_{\frac xt}$