How can I show $\lim_{t\to 0^{+}}\|S(t)-I\|\neq 0$ where $S(t)f(x)=e^{-t^2-2tx}f(x+t)$.. 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 $\displaystyle \|A\|=\sup_{\|x\|=1}\|Ax\|$.
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.
When $t-&gt;0$, $S(t)f(x)\sim e^{-2tx}.f(x)$

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

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

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

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