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

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?

• Even with the multi-variable substitution being true, your simplified desired equality is not equivalent, as the argument of $f$ should be modified. Oct 7 '18 at 11:54
• It should be $\int_{\mathbb{R}^n} f(x_0+\frac{y}{\alpha})\phi(y) dy= f(x_0)$. Oct 7 '18 at 12:02
• Oh I understand: If I do the substitution with the integral also having $f$, then the $x$ in $f$ gets substituted aswell. Thanks! Oct 7 '18 at 12:05
• Do you need additional help aside from that? Oct 7 '18 at 12:10
• I think I understand how to solve it: correctly using substitution in $\lim_{\alpha \to \infty} \int_{\mathbb{R}^n} f(x) \phi_{\alpha}(x - x_0) dx$ gives me $\lim_{\alpha \to \infty}$ of your function. If it is possible to insert the limit into the integral, I get $f(x_0)$ which is independent from the integral, so I can take it out as a constant factor and the rest of the integral is equal to 1. Is this correct? Then my only help needed is which theorem I can use to check if I can put the limit inside. Oct 7 '18 at 12:17

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 $$\alpha$$.
• Thank you, this means my problem should be solved! Oct 7 '18 at 13:46

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.