I am currently reading this paper about extending a smooth function and I'm a bit confused. Here is the context. I am trying to prove the following theorem:
Theorem. Let $f:\mathbb R^+ \rightarrow \mathbb R$ be a smooth function, and suppose the one-sided limits from above $\lim_{t \searrow 0} f^{(n)}(t)$ exist for all integers $n \geq 0$. Then there exists a smooth function $g: \mathbb R \rightarrow \mathbb R$ which agrees with $f$ on $\mathbb R^+$.
I understand the proof of the following lemma:
Lemma. There exist real sequences $(a_k)_{k=0}^\infty$ and $(b_k)_{k=0}^\infty$ such that $b_k < 0$ with $b_k \rightarrow - \infty$ as $k \rightarrow \infty$, $\sum_{k=0}^\infty |a_k||b_k|^n < \infty$ for any $n \geq 0$, and $\sum_{k=0}^\infty a_k(b_k)^n = 1$ for any $n \geq 0$.
The paper defines $g: \mathbb R \rightarrow \mathbb R$ via $$g(t) := \begin{cases} f(t) & \text{ if $t > 0$} \\ \lim_{t \searrow 0}f(t) & \text{ if $t = 0$} \\ \sum_{k=0}^\infty \phi(b_kt) f(b_kt) & \text{ if $t < 0$}, \end{cases}$$ where $\phi: \mathbb R \rightarrow \mathbb R$ is a smooth function with compact support which is equal to one on a neighbourhood of zero. The expression for $g(t)$ when $t < 0$ makes sense, because $b_k t \rightarrow \infty$ as $k \rightarrow \infty$, so the summation is finite.
Here is the part which I don't understand: the paper claims that since $\sum_{k=0}^\infty |a_k||b_k|^n < \infty$, it follows that the limits of $g^{(n)}(t)$ as $t \nearrow 0$ exist. I am stumped on why this is true. I'm guessing that the series $\sum_{k=0}^\infty \phi(b_kt) f(b_kt)$ converges uniformly so we can interchange the limit and the series, but I don't know how to show this. Could I get a suggestion?
I think you have a typo: $g(t)$ is defined by $g(t) = \sum_{k=0}^\infty a_k \phi(b_k t)f(b_k t)$ for $t<0$. With this let $K>0$ be such that $\phi$ has support contained in $[-K,K]$. Since $f$ is smooth it is bounded on $[-K,K]$, say by $M>0$ so $|\phi(b_k t)f(b_k t)| \le \|\phi\|_{\infty}M$ for all $t\le 0$ (by considering the cases $b_kt < -K$ and $b_k t \in [-K,0]$ separately).
Therefore defining $M_k = |a_k| \|\phi\|_{\infty} M$ we see that $\sum_{k=0}^\infty a_k \phi(b_k t)f(b_k t)$ converges uniformly on $(-\infty,0]$ by the Weierstrass $M$-test (as $\sum_{k=0}^\infty |a_k|<\infty$). Thus the interchange of limits is justified as you suspected.
The same strategy works with $g^{(n)}$, since the summand in this case will just be $a_k$ multiplied by some polynomial of $b_k,\phi^{(i)}(b_kt)$ and $f^{(i)}(b_kt)$ for $0\le i\le n$. The main difference is to instead take $M$ which bounds each of $f,f^{(1)},\ldots,f^{(n)}$ on $[-K,K]$.