Here is the problem:
Let $\phi(x) \in S$, where $S$ is the Schwartz class, such that $\displaystyle\dfrac{1}{\sqrt{2\pi}}\int_{\mathbb{R}} \phi=1$. Also, for some $N\in\mathbb{N}$, $\int_\mathbb{R}t^j\phi(t) dt=0$ for $j=1,2,...N-1$. Define $\phi_k(t)=k\phi(kt)$. Suppose that $f:\mathbb{R}\rightarrow \mathbb{C}$ is $N$ times differentiable with $f$ and $f^{(N)}$ both bounded. Show that there exists a constant $C$ such that
$$|f(x) \star \phi_k(x) -f(x)|\leq \frac{C}{k^N}$$
for all $x\in\mathbb{R}$ and $k\in\mathbb{N}$.
Here's what I've done so far:
Let $x\in\mathbb{R}$ and $N\in\mathbb{N}$.
I used Taylor's Theorem so that I can write $f$ as: $$f(t)=\sum_{j=0}^{N-1} \frac{f^{j}(x)}{j!}(t-x)^j +\frac{f^{(N)}(E)}{N!}(t-x)^n$$
for some $E$ between $x$ and $t$. Therefore,
$\displaystyle f(x) \star \phi_k(x) -f(x)= \int_{\mathbb{R}}f(t)\phi_k(t-x) dt= \left(\int_\mathbb{R}\sum\frac{f^{(j)}(x)}{j!}(t-x)^j\phi_k(t-x) dt +\int_{\mathbb{R}} \frac{f^{(N)}(E)}{N!}(x-t)^N\phi_k(t-x)\right)-f(x)$.
After evaluating the first term, the only thing that would be left is $\sqrt{2\pi}f(x)$.
And for the second term, I would be left with $\dfrac{f^{(N)}(E)}{N! k^N} \int u^n\phi(u)$ where $u=k(t-x)$.
I'm a bit stuck here. I can't seem to prove that the integrand above is bounded. Maybe I'm missing some facts about the Schwartz class.
Thank you in advance for responses.