Properly estimate a Schwartz function

Question

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 $|\alpha| \leq 2M$.

Answer 1

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}