Prove that for $\phi \in D$, $\lim_{n\to +\infty}\int_{\mathbb{R}}f_{n}(x)\phi (x)\,dx=1$

Question

Let defined sequence function :

$$f_{n}(x)=\begin{cases}1-\frac{|x|}{n},|x|<n\\0,\operatorname{otherwise}\end{cases}$$

Find

$\lim_{n\to +\infty}f_{n}(x)$

Then prove that for $\phi \in D$

$\displaystyle\lim_{n\to +\infty}\int_{\mathbb{R}}f_{n}(x)\phi (x)dx$=1$

My solution for first question

Take limit of $f_{n}$ so

$\lim_{n\to +\infty}\left(1-\frac{|x|}{n}\right)=1$

But $|x|<+\infty$

So what's the limit of $f_{n}(x)$ ?

Answer 1

Fix $x \in \mathbb{R}$ and notice that for all $n \in \mathbb{N}$ such that $|x| > n$ we have $$f_n(x) = 1- \frac{|x|}n \xrightarrow{n\to\infty} 1$$ so $\lim_{n\to\infty} f_n(x) = 1$.

Your second claim seems to be that for all $\phi \in \mathcal{D}(\mathbb{R})$ holds $$\lim_{n\to\infty}\int_{\mathbb{R}}f_n(x)\phi(x)\,dx = 1$$ but this is false e.g. for $\phi \equiv 0$.