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For every Schwartz function $\varphi \in S(\mathbb{R}^{n})$ there exists a constant $c_{\beta, k}$ such that one can estimate the Schwartz function by $$|\partial^{\beta}\varphi(x)| \leq \frac{c_{\beta, k}}{(1 + |x|^{2})^{k}} , $$ where $\beta \in \mathbb{N^{n}}$ is a multi-index and $k \in \mathbb{N}$. However, when one is dealing with tempered distributions and one wants to show that such a distribution is continuous one often chooses to estimate the test function - let's call it $\phi \in S(\mathbb{R}^{n})$ - as follows: $$|\phi(x)| \leq C_{M, n}\frac{\sum_{|\alpha| \leq2M} \left\lVert \phi\right\rVert_{\alpha,0}}{(1 + \left\lVert x\right\rVert^{2})^{M}} , $$ where $\left\lVert \phi\right\rVert_{\alpha,\beta} = \sup_{x \in \mathbb{R^{n}}}|x^{\alpha}\partial^{\beta}\phi(x)|$ is the "Alpha-Beta" norm of the Schwartz space and $\alpha, \beta$ are multi-indices, $M \in \mathbb{N}$ is arbitrary and the constant $C_{M,n} > 0$ is chosen such that the estimate holds ($n$ is the dimension of the Schwartz space). How come the numerator of the second inequality contains a summation over certain norms? I don't understand why one can't simply use the estimation that only involves one constant in the numerator, such as in the first inequality. It isn't either clear to me why the summation is only carried out as long as $|\alpha| \leq 2M$.

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Observe \begin{align} \left|(1+|x|^2)^M\phi(x)\right| \leq&\ \left|(1+M|x|^2+\ldots+|x|^{2M})\phi(x) \right|\\ \leq&\ |\phi(x)|+M|x|^2|\phi(x)|+\ldots +|x|^{2M}|\phi(x)|. \end{align} Next, note that \begin{align} |x|^k|\phi(x)| \leq&\ C_n\sum_{|\alpha|=k}|x^\alpha \phi(x)|. \end{align} Finally, it follows \begin{align} |\phi(x)|+M|x|^2|\phi(x)|+\ldots +|x|^{2M}|\phi(x)| \leq C_{M, n}\sum_{k\leq 2M}\sum_{|\alpha|=k}|x^\alpha\phi(x)|. \end{align}

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Depends what one is trying to do. If you have a candidate for a temperate distribution say $T(\phi)=\int f(x)\phi(x) d^n x$ and you want to show it is indeed a distribution you have to establish continuity, i.e., an inequality of the form $|T(\phi)|\le C \rho(\phi)$ where $\rho$ is a continuous seminorm on Schwartz space, for example the numerator of the 2nd inequality. One would typically use this 2nd inequality to bound $|\phi(x)|$ in the integral, whereas $M$ would have to be chosen large enough so $\int |f(x)|(1+|x|^2)^{-M}d^nx\ <\infty$ which will give you the constant $C$ in the continuity statement.

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