Is translation continouos in Schwartz space?

Question

It is true that translation is continuous in Schwartz Space {S}($\mathbb{R}$) with its topology?, in other words, I'm trying to prove that if $\phi \in {S}\left(\mathbb{R}\right)$ then the function $\phi(\cdot-y)$ converges to $\phi$ in $S\left(\mathbb{R}\right)$. So I have to prove that for all $\alpha, \beta \in \mathbb{N}$. \begin{equation*} \lim _{ y\rightarrow 0 }{ { \left\| { T }_{ y }\left( \phi \right) -\phi \right\| }_{ \alpha ,\beta } } =0. \end{equation*} This is my attempt. It is known that fourier transformation is continuous in Schwartz,so it is enough to prove my idea that the fourier transformation of $T_y(\phi)$ will converge to $\phi$ in Schwartz. How \begin{equation*} \widehat { { T }_{ y }\left( \phi \right) } \left( \xi \right) =e^{ -i\xi y }\widehat { \phi } \left( \xi \right), \end{equation*} I have to prove that \begin{equation*} \lim _{ y\rightarrow 0 }{ { \left\| e^{ -i\xi y }\widehat { \phi } \left( \xi \right) -\widehat { \phi } \left( \xi \right) \right\| }_{ \alpha ,\beta } } =0. \end{equation*} Thus, \begin{align} \nonumber\left| { \xi }^{ \alpha }{ \left( e^{ -i\xi y }\widehat { \phi } \left( \xi \right) -\widehat { \phi } \left( \xi \right) \right) }^{ \left( \beta \right) } \right|& \le \sum _{ j=0 }^{ \beta }{ \left| { \left( e^{ -i\xi y }-1 \right) }^{ \left( j \right) } \right| } \left| { \xi }^{ \alpha }{ \widehat { \phi } \left( \xi \right) }^{ \left( \beta -j \right) } \right| \\\nonumber &\le C\sum _{ j=0 }^{ \beta }{ \left| { \left( e^{ -i\xi y }-1 \right) }^{ \left( j \right) } \right| } \\ \nonumber&\le C{ \left( \sum _{ j=1 }^{ \beta }{ \left| { \left( -iy \right) }^{ j } \right| } +\left| 1-e^{ -i\xi y } \right| \right) } \end{align} Then, the function $\left| 1-e^{ -i\xi y } \right| $ is continous and periodic, so its maximun exists,and that follows

\begin{align} \nonumber\sup_{x\in \mathbb{R}}\left| { \xi }^{ \alpha }{ \left( e^{ -i\xi y }\widehat { \phi } \left( \xi \right) -\widehat { \phi } \left( \xi \right) \right) }^{ \left( \beta \right) } \right|&\le C{ \left( \sum _{ j=1 }^{ \beta }{ \left| { \left( -iy \right) }^{ j } \right| } +\left| 1-e^{ -iz y } \right| \right) } \end{align} where z is where the function gets its maximum, then i take the limit when y->0, and the prove is finished.

Answer 1

I'll write $f$ for $\phi.$ Let $m,n\in \{0,1,2\dots \}.$ We want to show

$$\sup_x |x|^m|D^n(f(x-y)-f(x))| \to 0$$

as $y\to 0.$ Now $D^n (f(x-y)-f(x))$ $ = D^nf(x-y)-D^nf(x).$ And we know $D^nf \in \mathcal S.$ So it's enough to show that if $g\in \mathcal S,$ then

$$\tag 1 \sup_x |x|^m|g(x-y)-g(x))| \to 0.$$

By the MVT we have $g(x-y)-g(x) = g'(c)(-y)$ for some $c$ between $x-y$ and $x.$ Thus

$$|g(x-y)-g(x))|\le \left (\sup_{z\in B(x,|y|)}|g'|\right)|y|.$$

So we'll be done if we show

$$|x|^m\left (\sup_{z\in B(x,|y|)}|g'|\right)$$

is bounded independent of $x.$ Fix $x,$ and let $z\in B(x,|y|).$ Then $z=x+t,$ where $|t|\le |y|.$ We have

$$|x|^m|g'(x+t)| \le (|x+t| + |t|)^m|g'(x+t)|$$ $$ \le (2^m|x+t|^m + 2^m|y|^m)|g'(x+t)|.$$

Because $g \in \mathcal S,$ $|x+t|^m|g'(x+t)|$ is bounded independent of $x,$ and so is $|y|^m|g'(x+t)|.$ We thus have desired boundedness, giving the desired result.