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?

  • 1
    $\begingroup$ Even with the multi-variable substitution being true, your simplified desired equality is not equivalent, as the argument of $f$ should be modified. $\endgroup$ Oct 7 '18 at 11:54
  • 1
    $\begingroup$ It should be $\int_{\mathbb{R}^n} f(x_0+\frac{y}{\alpha})\phi(y) dy= f(x_0)$. $\endgroup$ Oct 7 '18 at 12:02
  • $\begingroup$ Oh I understand: If I do the substitution with the integral also having $f$, then the $x$ in $f$ gets substituted aswell. Thanks! $\endgroup$
    – Geckabor
    Oct 7 '18 at 12:05
  • 1
    $\begingroup$ Do you need additional help aside from that? $\endgroup$ Oct 7 '18 at 12:10
  • $\begingroup$ 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. $\endgroup$
    – Geckabor
    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 $\alpha$.
  • $\begingroup$ Thank you, this means my problem should be solved! $\endgroup$
    – Geckabor
    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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.