How prove it such that the sequence $\{L_{f}(\phi_{j})\}$ does not tend to zero. Let $f(x)=e^{x^2}$, For $\phi\in D(R)$,define $L_{f}(\phi)=\int f\phi dx$;
construct a sequence of function $\{\phi_{j}\}$ in $D$ that tends to zero in $S$ but such that the sequence $\{L_{f}(\phi_{j})\}$ does not tend to zero.
where $S$ is meaning Schwartz class  can see:http://www.math.mcgill.ca/gantumur/math581w12/downloads/Lecture12.pdf
the book:
Hint: Note that $L_{f}(\phi_{j})$ does tend to zero if $\{\phi_{j})$ tends to zero in $D$.so you need a sequence of functions $\{\phi_{j}\}$ converging to zero in $S$ but not in $D$.
How prove it this problem ,Thank you 
 A: I hope the following example works.
Take a function $\phi\in\mathcal D$ such that $\phi\geq 0$, $\phi$ is supported on $[0,3]$ and $\phi(x)\equiv 1$ on $[1,2]$. Then define $\phi_j(x)=\varepsilon_j\,  \phi(x-j)$, where $\varepsilon_j=e^{-j}$.
Let us first show that $\phi_j\to 0$ in $\mathcal S$. For any $\psi\in\mathcal S$ and $k,l\in\mathbb N$, put $$\Vert\psi\Vert_{k,l}=\sup\,\{ \vert x^k\vert \,\vert \psi^{(l)}(x)\vert;\; x\in\mathbb R\}$$
Note that $\phi_j^{(l)}(x)=\varepsilon_j \phi^{(l)} (x-j)$ and that $\phi_j$ is supported on $[j,j+3]$. So for any $k\in\mathbb N$ we have 
$$\Vert \phi_j \Vert_{k,l}\leq\varepsilon_j (j+3)^k \Vert \phi^{(l)}\Vert_\infty
= e^{-j} (j+3)^k \Vert \phi^{(l)}\Vert_\infty\, ,$$
and hence $\Vert \phi_{j}\Vert_{k,l}\to 0$ as $j\to\infty$. This shows that $\phi_j\to 0$ in $\mathcal S$.
On the other hand, we have $\phi(x-j)\equiv 1$ on $[j+1,j+2]$. Since $\phi\geq 0$ everywhere it follows that 
$$L_f(\phi_j)\geq \varepsilon_j\,\int_{j}^{j+1} e^{x^2} dx \geq \varepsilon_j\, e^{(j+1)^2}=e^{-j+(j+1)^2}\, ,$$
and hence $L_f(\phi_j)\to +\infty$.
