How can I show $\lim_{t\to 0^{+}}\|S(t)-I\|\neq 0$ where $S(t)f(x)=e^{-t^2-2tx}f(x+t)$..

Question

I need some help with the following:

For every $t\in [0, \infty)$ let $S(t):C_0([0, \infty))\rightarrow C_0([0, \infty))$ be the bounded linear operator given by, $$S(t)f(x)=e^{-t^2-2tx}f(x+t)),$$ where $\displaystyle C_0([0, \infty))=\{f\in C^\infty([0, \infty), \mathbb R): \lim_{x\to \infty}f(x)=0\}$. How to show $$\lim_{t\to 0^{+}}\|S(t)-I\|\neq 0.$$ Recall the norm of a bounded linear operator is $\displaystyle \|A\|=\sup_{\|x\|=1}\|Ax\|$.

Answer 1

When $t->0$, $S(t)f(x)\sim e^{-2tx}.f(x)$

So $\forall x \in [0, \infty[, \lim_{t->0^+}{\|S(t)-I\|}=0 $ because $e^{-2tx} = 1$

But when $x->\infty$, $e^{-2tx}$ is bounded (between 0 and 1) and $f(x)->0$ so :

$$\lim_{t->0^+}{S(t)f(t)}=\lim_{x->\infty} e^{-2tx}.f(x) = \lim_{x->\infty}f(x)=0 \neq I$$

So $lim_{t->0^+}{[\sup_{x\in[0,\infty[}{\|(S(t)-I)(f(x))\|}]}\neq 0 $

Answer 2

Notice that if we choose, for each $t$, $f(x)=e^{-x/t}$.

$S(t)f(0)=e^{-t^2-1}$, so $S(t)f(0)- f(0) = e^{-t^2-1} -1 $

Hence, by definition of sup, $||S(t)-I|| \geq 1 - e^{-t^2-1}$. Then we are done.

This is a an example of pointwise convergence not implying uniform convergence.