What is wrong with my application of Lebesgue Dominated Convergence Theorem in these two examples?

Question

Background

I seem to be having issues recognizing valid bounding functions when applying the Lebesgue Dominated convergence theorem. Here are two examples I did that I do not think are justified.

Example 1: Show that $$\lim_{n\to\infty} \int_0^\infty ne^{-nx}\sin(1/x) \ dx$$ exists and determine its value.

Solution: Let $u=nx$. Then we may rewrite the integral as $$\int_0^\infty e^{-u}\sin\left(\frac{n}{u}\right) \ du.$$ Note that $$\bigg\lvert e^{-u}\sin\left(\frac{n}{u}\right)\bigg\rvert\leq e^{-u},$$ and since $e^{-u}$ is integrable, the LDCT applies. The limit is then $0$.

Example 2: Let $g:\mathbb{R} \rightarrow \mathbb{R}$ be integrable and let $f:\mathbb{R} \rightarrow \mathbb{R}$ be bounded, measurable, and continuous at $1$. prove that $$\lim_{n\to\infty}\int_{-n}^nf\left(1+\frac{x}{n^2}\right)g(x) \ dx$$ exists and determine its value.

Solution: $f$ is bounded so for all $x$,$|f|\leq K$ for some $k \in \mathbb{R}$. Thus $$\bigg\lvert f\left(1+\frac{x}{n^2}\right)g(x) \chi_{[-n,n]}\bigg\rvert\leq k|g(x)|.$$ Since $k|g(x)|$ is integrable, we again have by the LDCT that the integral converges. The limiting value is $\int_\mathbb{R}f(1)g(x) \ dx.$

My Question

I am not really looking for the solution to these problems. Rather I wish to know how I am applying the LDCT wrong in both of these (I am fairly sure that I am applying it wrong).

Answer 1

The second one seems fine to me.

For the first, you simply hid the dependency on $n$.

That is, it's fine to set $u = nx$ inside the limit; for you get that for every fixed $n$, your integrand is bounded by $e^{-u}$. But anyway your $u$ depends on $n$; so what you've shown is that the $n$-th integrand can be bound by a function that depends on $n$, and this is not what the LDCT requires.

Answer 2

In Example 1, after your substitution, the integral should be $$ \lim_{n\to\infty}\int_0^\infty e^{-u}\sin\left(\frac nu\right)\,\mathrm{d}u $$ Indeed, the integrand is dominated by $e^{-u}$, but it doesn't converge pointwise to $0$.

Let's try integrating by parts, then substitute $u=nx$ $$ \begin{align} \int_0^\infty ne^{-nx}\sin\left(\frac1x\right)\,\mathrm{d}x &=\int_0^\infty nx^2e^{-nx}\,\mathrm{d}\cos\left(\frac1x\right)\\ &=\int_0^\infty\cos\left(\frac1x\right)(n^2x^2-2nx)e^{-nx}\,\mathrm{d}x\\ &=\int_0^\infty\frac1n\cos\left(\frac nu\right)(u^2-2u)e^{-u}\,\mathrm{d}u \end{align} $$ Now, the integrand is bounded by $|u^2-2u|\,e^{-u}$, which is integrable, and the integrand converges pointwise to $0$.

In fact, since $\int_0^\infty|u^2-2u|\,e^{-u}\,\mathrm{d}u=\frac8{e^2}$, we have $$ \left|\,\int_0^\infty ne^{-nx}\sin\left(\frac1x\right)\,\mathrm{d}x\,\right|\le\frac8{ne^2} $$

Answer 3

The mistake in the first problem is very subtle: you first note that $f_n \to 0$ pointwisely; after the change of variables you get some new functions such that $|g_n| \le \Bbb e ^{-u}$, but how do you connect the $f$s and the $g$s? Plus, $g_n \not\to 0$.

The second one is fine.