Prove that for $\phi\in D$ : $\lim_{n\to +\infty}\int_{\mathbb{R}}f_n(x)\phi (x)dx=\delta$

Question

Let defined the sequence of function for $n\in\mathbb{N^{*}}$ :

$$f_{n}(x)=\begin{cases}-n,|x|<\frac{1}{2n}\\2n,\frac{1}{2n}<|x|<\frac{1}{n}\\0,|x|>\frac{1}{n}\end{cases}$$

Then evaluate the following limit :

$a)$ $\lim_{n\to +\infty}f_n(x)=?$

$b)$ for $\phi\in D$ prove that :

$\lim_{n\to +\infty}\int_{R}f_n(x)\phi (x)dx=\delta$

I can't started in this limits I don't have any Hint's or ideas for this type sequence function

I have already see your solution !

Answer 1

Hints:

(a) Fix $x \in \mathbb{R}$ and notice that for all $n \in \mathbb{N}$ such that $|x| > \frac1n$ we have $f_n(x) = 0$ so $f_n(x) \to 0$.

(b) I'm assuming that $D$ is actually the space $C^\infty_c(\mathbb{R})$. Let $\varepsilon > 0$. Since $\phi$ is continuous at $0$ there exists $n_0 \in \mathbb{N}$ such that $|x| < \frac1{n_0}$ implies $|\phi(0) - \phi(x)| < \frac\varepsilon3$.

Notice that $\int_\mathbb{R}f_n = 1$ and $\int_\mathbb{R}|f_n| = 3$ so for all $n \ge n_0$ we have $$\left|\phi(0) - \int_\mathbb{R}f_n(x)\phi(x)\,dx\right| = \left| \int_\mathbb{R}f_n(x)(\phi(0)-\phi(x))\,dx\right| \le \int_{-\frac1n}^{\frac1n}|f_n(x)||\phi(0) - \phi(x)| < \varepsilon$$