Show that $\lim_{\alpha \to \infty} \int_{\mathbb{R}^n} f(x) \phi_{\alpha}(x - x_0) dx = f(x_0)$

Question

Given is a integrable function $\phi: \mathbb{R}^n \mapsto \mathbb{R}$ with the property $\int_{\mathbb{R}^n} \phi dx = 1$. We define for $\alpha > 0$ the re-scaled function $\phi_\alpha(x):= \alpha^n \phi(\alpha x) $. Now I should show that for every continuous and bounded function $f: \mathbb{R}^n \mapsto \mathbb{R}$ and for every $x_0 \in \mathbb{R}^n$ the following holds:

$\lim_{\alpha \to \infty} \int_{\mathbb{R}^n} f(x) \phi_{\alpha}(x - x_0) dx = f(x_0)$

So I've done some research and found that the properties of $\phi$ are very similar to something called the dirac-delta function $\delta$. One property of $\delta$ is that $|\alpha| \delta(\alpha x) = \delta(x)$.

I will now prove that something similar holds for $\phi$, which means I will show that $\alpha^n \phi(\alpha x) = \phi(x)$ ($\alpha$ already is $>0$ so I can leave out the absolute value).

Using multivariable substitution ($u = \alpha x$):

$\int_{\mathbb{R}^n} \alpha^n \phi(\alpha x) dx = \int_{\mathbb{R}^n} \alpha^n \phi(u) \frac{1}{\alpha^n} du = \int_{\mathbb{R}^n} \phi(u)du = 1$

If this is correct, my problem becomes easier and I only have to show that

$\int_{\mathbb{R}^n} f(x)\phi(x-x_0) = f(x_0)$

I am not sure how to continue from this point on, but my guess would be that using the property of $f$ being bounded might prove useful. Also please correct me on mistakes I have done thus far.

EDIT: Thanks to help in the comments I maybe figured out how to correct my mistakes and continue:

Using multivariable substitution ($u = \alpha x - \alpha x_0$):

$\lim_{\alpha \to \infty} \int_{\mathbb{R}^n} f(x) \alpha^n \phi(\alpha x) dx = \lim_{\alpha \to \infty} \int_{\mathbb{R}^n} f(x_0 +\frac{u}{\alpha}) \phi(u) du$

Now if it is possible to get $\lim$ inside of the integral we have:

$\int_{\mathbb{R}^n} \lim_{\alpha \to \infty} f(x_0 +\frac{u}{\alpha}) \phi(u) du = \int_{\mathbb{R}^n} f(x_0) \phi(u) du = f(x_0)\int_{\mathbb{R}^n} \phi(u) du = f(x_0)$

Is this correct? If yes, which theorem can I use to get the limit inside the integral?

Answer 1

You can use the Dominated Convergence Thoerem. Let's check that the hypothesis are satisfied.

1. By continuity, $\lim_{\alpha\to\infty} f\Bigl(x_0 +\dfrac{u}{\alpha}\Bigr)\, \phi(u) =f(x_0)\,\phi(u)$ for all $u$.
2. Let $M$ be a bound for $f$. Then $\Bigl|f\Bigl(x_0 +\dfrac{u}{\alpha}\Bigr)\, \phi(u)\Bigr|\le M\,|\phi(u)|$, and $M\,|\phi(u)|$ is integrable and independent of $\alpha$.

Answer 2

I'm assuming that by integrable function, you mean that:

$\int_{\mathbb{R}^n}\phi(x)dx= \underset{R\rightarrow \infty}{\lim} \int_{ \{ \Vert x\Vert\leq R \} }\phi(x)dx$.

Then proving that $f(x)\cdot \phi_{\alpha}(x-x_0)$ is integrable is to show that:

$\underset{R\rightarrow \infty}{\lim} \int_{ \{ \Vert x\Vert\leq R \} } f(x) \phi_\alpha(x-x_0)dx=\int_{\mathbb{R}^n} f(x) \phi_{\alpha}(x - x_0) dx$

So by integrability of $f(x)\cdot \phi_{\alpha}(x-x_0)$ you could pass to the limit. To show intgrablitiy it would be useful to remember that $f$ is bounded.