Consider the following function:

$\delta_n(t)$ = \begin{cases} n^{2}(t+1/n) &\quad\ t\in [-1/n, 0]\\ -n^{2}(t-1/n) &\quad\ t\in [0, 1/n]\\ 0 &\quad\text{otherwise} \\ \end{cases}

which is ''triangle'' of height $n$ and base $[-1/n, 1/n]$.

Now, I have to prove that for each bounded and continuous function $\varphi(t)$, we have that:

$$\lim_{n\to\infty}\int_{-\infty}^{+\infty}\delta_n(t)\varphi(t)dt = \varphi(0)$$

I cannot mention the Delta Dirac function since it has not been introduced.

So far, I have proved that:

$$\int_{-\infty}^{+\infty}\delta_n(t)dt = 1 $$

And then I was thinking about adding and subtracting $\varphi(0)$ inside the integral obtaining:

$$\int_{-\infty}^{+\infty}\delta_n(t)(\varphi(t)-\varphi(0)+\varphi(0))dt = \int_{-\infty}^{+\infty}\delta_n(t)\varphi(0)dt + \int_{-\infty}^{+\infty}\delta_n(t)(\varphi(t)-\varphi(0))dt $$

Where with the first term I take out $\varphi(0)$ and then compute the limit and it is obviously equal to $\varphi(0)$. But I have to prove that the result of the second is 0. Any suggestions?

  • $\begingroup$ Use the $\epsilon, \delta$ definition of "continuous" $\endgroup$ – GEdgar Jun 23 '18 at 10:50

Given $\varepsilon > 0$, there is some $\rho> 0 $ such that for every $ t \in (-\rho,\rho)$ we have $|\varphi(t) - \varphi(0)| < \varepsilon$, since $\varphi$ is continuous at the origin. Now for any $n \in \mathbb{N}$ with $n > \frac{1}{\rho} \Leftrightarrow \frac{1}{n}<\rho$ we find \begin{align} \left| ~\int \limits_{-\infty}^\infty \delta_n (t) (\varphi(t) - \varphi(0)) \, \mathrm{d} t ~\right| &= \left| ~\int \limits_{-1/n}^{1/n} \delta_n (t) (\varphi(t) - \varphi(0)) \, \mathrm{d} t ~\right| \\ &\leq \int \limits_{-1/n}^{1/n} \delta_n (t) |\varphi(t) - \varphi(0)| \, \mathrm{d} t \\ &< \varepsilon \int \limits_{-1/n}^{1/n} \delta_n (t) \, \mathrm{d} t \, . \end{align} Now just use your result for the integral of $\delta_n$ and the definition of limits to conclude that the second term vanishes as $n \rightarrow \infty$.


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