Theorem:
$$\hat{(D^{\alpha}u)} = (iy)^\alpha \hat{u}$$
So, in order to solve
$$-\Delta u + u = f$$
Evan's book applies the Fourier transform on both sides and ends up with
$$(1+|y|^2)\hat{u}(y) = \hat{f}(y)$$
I'm trying to understand this, because
$$-\Delta u + u = -\left(\frac{\partial^2 u}{\partial y_1^2}+\cdots + \frac{\partial^2 u}{\partial y_n^2}\right)+u$$
or you can view each $\frac{\partial^2 u}{\partial y_1^2}$ as $D^{\alpha}u$ such that $|\alpha|=2$, so for each term $D^{\alpha}$ we should get $\hat{D^\alpha u} = (iy)^{\alpha}\hat u$ where $|\alpha|=2$, and we should sum them all to get
$$-((iy)^{\alpha_1}+\cdots (iy)^{\alpha_n})+\hat{u} = \hat{f}$$
How does that transforms into $(1+|y|^2)\hat{u}(y) = \hat{f}(y)$?
If $y=(y_1,\dots,y_n)$ and $\alpha=(\alpha_1,\dots,\alpha_n)$, then $$ y^\alpha=y_1^{\alpha_1}\dots y_n^{\alpha_n}\quad\text{and}\quad|\alpha|=\alpha_1+\dots\alpha_n. $$ We have $$ \frac{\partial^2 u}{\partial x_k^2}=D^\alpha u\quad\text{with}\quad \alpha=(0,\dots,0,\underbrace{2}_{k\text{ position}},0,\dots,0) $$ Then $$ \widehat{\frac{\partial^2 u}{\partial x_k^2}}(y)=-y_k^2\,\hat u(y) $$ and $$ -\widehat{\Delta u}(y)=\sum_{k=1}^n\bigl(y_k^2\,\hat u(y)\bigr)=\Bigl(\sum_{k=1}^ny_k^2\Bigr)\,\hat u(y)=|y^2|\,\hat u(y). $$