I need help with this problem:

Suppose that $f$ is a Lebesgue integrable function defined on $\mathbb{R}^d$. Prove that $$\int_{\mathbb{R}^d} f(x)e^{-\frac{1}{n}\|x\|}dx \to \int_{\mathbb{R}^d} f(x)dx.$$

If I define $f_n(x) = f(x)e^{-\frac{1}{n}||x||}$, then I am having trouble finding a suitable integrable $\phi(x)$ such that $|f_n(x)| \le \phi(x)$.

Thanks in advance!


$0\leq e^{-\frac{1}{n}\|x\|}\leq1$ so that $|f_n|=|f|e^{-\frac{1}{n}\|x\|}\leq |f|$ and then you can use DCT.

Edit: We can also use MCT: First we have $$\left|\int_{\mathbb R^d}f(x)e^{-\frac{1}{n}\|x\|}\,dx -\int_{\mathbb{R}^d} f(x)\,dx\right|\leq\int_{\mathbb{R}^d}|f(x)|(1-e^{-\frac{1}{n}\|x\|})\,dx.$$ Since $|f(x)|e^{-\frac{1}{n}\|x\|}\nearrow|f(x)|$ for all $x\in\mathbb R^d$, an application of MCT to $|f(x)|e^{-\frac{1}{n}\|x\|}$ gives that $$\int_{\mathbb R^d}|f(x)|e^{-\frac{1}{n}\|x\|}\,dx\nearrow\int_{\mathbb R^d}|f(x)|\,dx$$ and therefore $$\int_{\mathbb{R}^d}|f(x)|(1-e^{-\frac{1}{n}\|x\|})\,dx\to0\ \ \text{as } n\to\infty,$$ so $$\int_{\mathbb R^d}f(x)e^{-\frac{1}{n}\|x\|}\,dx \to\int_{\mathbb{R}^d} f(x)\,dx\ \ \text{as } n\to\infty.$$ But unlike the other answer, my opinion is that it is easier to use DCT than MCT.


The easiest way (in my opinion) is not to use dominated convergence, but rather monotone convergence.

  • 6
    $\begingroup$ How exactly? We don't have $f \ge 0$. $\endgroup$ – mechanodroid Jul 16 at 10:32
  • $\begingroup$ @mechanodroid See my edited answer. $\endgroup$ – Feng Shao Jul 16 at 23:44

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.