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The theorem I'm trying to understand is the following:

For $i=0,1,...$ let $f_i\in C^\infty_c(K)$ where $K$ is a compact subset of $\mathbb{R}^n$, and let $I$ be a compact neighborhood of $0$ in $\mathbb{R}$. Then one can find $f \in C^\infty_c(K\times I)$ such that $\partial^i f(x,t)/\partial t^i = f_i(x),\;\; t=0, i=0,1,...$

Here $C^\infty_c(K)$ is the set of infinitely continuously differentiable functions on K with compact support.

For an example of such a function we have $\Theta(t)=e^{-\frac{1}{1-t^2}}\;$ for $t>0$ and $\Theta(t)=0\;$ for $t\leq 0$.

Now to the proof of the theorem:

Choose $h\in C^\infty_c(I)$ so that $\text{d}^{i}(h(t)-1)/\text{d}t^i = 0$ when $t=0$ for $i=0,1,...$

For example, if $(-\epsilon,\epsilon)\in I$ we can take $\Theta$ with support in $(0,\epsilon)$ and $\int \Theta(t)dt=1$, and let $h$ be the solution of $h'(t) = \Theta(-t) - \Theta(t)$ with support in $I$.

I understand how I can choose $\Theta$ so that its support is in $(0,\epsilon)$ and it integrates to $1$. However, what I don't get is why $h$ has to be the solution of said ODE in order for $\text{d}^{i}(h(t)-1)/\text{d}t^i = 0$ to hold

also: isn't $\Theta$ symmetrical such that $\Theta(-t)=\Theta(t)$?

On with the proof:

Now, $h_i(x,t) := h(t/\epsilon_i)t^if_i(x)/i!$ is in $C^\infty_c(K\times I)$, and $|\partial^\alpha h_i(x,t)|\leq 2^i$ if $|\alpha|\leq j-1$, provided that $\epsilon_i$ is sufficiently small.

How do I see that $|\partial^\alpha h_i(x,t)|\leq 2^i$?

I left out the rest of the proof, since I want to first try to understand it up to this point.

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