Rudin's Functional Analysis Theorem 7.23 At the end of (b) of Theorem 7.23 in Rudin's Functional Analysis, it says

"Since (1) holds for $z \in \mathbb{R}^n$ (by the choice of $u$), Lemma 7.21 completes the proof of (b)."

Here, (1) is $\mathbf{f(z) = u(e_{-z})}$, where $u \in \mathscr{D}'(\mathbb{R}^n)$. The latter part of the statement is clear, so the only thing I have to understand is why (1) holds for $z \in \mathbb{R}^n$.
The condition for $f$ is that $\vert f(x) \vert \le \gamma(1 + \vert x \vert)^N $ and the book goes on to say

"The restriction of $f$ to $\mathbb{R}^n$ is therefore in $\mathscr{S}_n'$ and is the Fourier transform of some tempered distribution." $(*)$

However, I really don't know why (1) holds for $z \in \mathbb{R}^n$ only based on the above statement $(*)$, especially provided that $e_{-z}(t) = e^{-iz \cdot t}$ is not in $\mathscr{S}_n'$ and $f$ is not in $L^1(\mathbb{R}^n)$. Maybe the approximation works, but I am still struggling to show (1).
Any help would be appreciated!
 A: Hint: If $\phi \in \mathcal S_n$ then $\hat {\phi} (y)=\int \phi (x) e_{-y} (x) dx$. Apply $u$ to get $\int f \phi=\int u(e_{-y}) \phi$ for all $\phi \in \mathcal S_n$.
A: Let's show $f(x) = u(e_{-x})$. Even though $u \in \mathscr{D}(\mathbb{R}^n)$, we can extend it to $C^\infty(\mathbb{R}^n)$. To do that, it suffices to show $$\int_{\mathbb{R}^n} f(x)\phi(x)\,dx = \int_{\mathbb{R}^n} u(e_{-x})\phi(x)\,dx $$ for all $\phi \in \mathscr{D}(\mathbb{R}^n)$.
We first note that $$\int_{\mathbb{R}^n} f(x)\phi(x)\,dx = \langle f, \phi\rangle = \hat{u}(\phi) = u(\hat{\phi})$$.
Since $\hat{\phi}\in \mathscr{S}(\mathbb{R}^n)$, $\hat{\phi}(y) = \int_{K} \phi(x)e^{-ix\cdot y}\,dx $, where $K$ is the compact support of $\phi$. Now, let's define $g: K \subset \mathbb{R}^n \to C^\infty(\mathbb{R}^n)$ with $g(x) = \phi(x)e_{-x} $.
Then, since $g(x)$ is continuous, $K$ is a compact set in $\mathbb{R}^n$ and $C^\infty(\mathbb{R}^n)$ is a Frechet space, $\int_K \phi(x)e_{-x}\,dx$ is well-defined and  $\Lambda(\int_K \phi(x)e_{-x}\,dx) = \int_K \Lambda(\phi(x)e_{-x})\,dx$ by Thm 3.27.
If we let $z = \int_K \phi(x)e_{-x}\,dx$, then $z(y)$ is a continuous linear functional and $$z(y) = \int_K \phi(x)e_{-x}(y)\,dx = \int_K \phi(x)e^{-ix\cdot y}\,dx = \hat{\phi}(y).$$ Therefore, $z = \hat{\phi}$ and $$u(\hat{\phi}) = u(z) = \int_K u(\phi(x)e_{-x})\,dx = \int_K \phi(x)u(e_{-x})\,dx$$.
