Let $\{f_n\}^\infty_{n=1}$ be a sequence of continuous real valued functions defined on $\mathbb{R}$

Question

Let $\{f_n\}^\infty_{n=1}$ be a sequence of continuous real valued functions defined on $\mathbb{R}$ which converges pointwise to a continuous real valued function $f$. d Which of the following statements are true?

a. If $0\le f_n \le f$ for all $n\in \mathbb{N}$, then

$$\lim_{n\to \infty} \int_\infty^{-\infty} f_n(t)\,dt=\int_\infty^{-\infty } f(t) \, dt$$

b. If $|f_n(t)|\le |\sin t|$ for all $t\in \mathbb{R}$ and for all $n\in \mathbb{N}$, then $$\lim_{n\to \infty} \int_\infty^{-\infty } f_n(t) \, dt=\int_\infty ^{-\infty } f(t) \, dt$$

c. If $|f_n(t)|\le e^t$ for all $t\in \mathbb{R}$ and for all $n\in \mathbb{N}$, then for all $a,b \in \mathbb{R}$ . $a<b$

$$\lim_{n\to \infty} \int_b^a f_n(t) \, dt=\int_b^a f(t) \, dt$$

since sequence of continuous real valued functions defined on $\mathbb{R}$ which converges pointwise to a continuous real valued function $f$ so 1 is true (but ia m not sure) can you hlep me with other options too..thank you

Answer 1

For case b., consider the sequence $(f_n)$ where $f_n(t) = \vert \sin t \vert$ for $t \in [n \pi, (n+1)\pi]$ and vanishes elsewhere. $(f_n)$ converges pointwise to the always vanishing function.

However, you have for all $n \in \mathbb N$

$$2 = \int_{-\infty}^\infty f_n(t) \ dt \neq \int_{-\infty}^\infty 0 \ dt =0.$$

Cases a. and c. are consequences of Lebesgue dominated convergence theorem providing $\int_{-\infty}^\infty f(t) \ dt$ converges for case a. The theorem doesn’t apply to case b. as $\int_{-\infty}^\infty \vert \sin t \vert \ dt$ diverges.

Answer 2

(a) and (c) are true. (b) is wrong.

- Proof (a). By Fatou's Lemma one has: \begin{align} \int_\mathbb{R} f(x)\,dx \leq \liminf_{n\to\infty} \int_\mathbb{R} f_n(x)\,dx \end{align} Moreover we have by monotonicity of the integral: \begin{align} \limsup_{n\to\infty}\int_\mathbb{R} f_n(x)\,dx \leq \int_\mathbb{R} f(x)\,dx \end{align} Hence: \begin{align} \limsup_{n\to\infty}\int_\mathbb{R} f_n(x)\,dx \leq \int_\mathbb{R} f(x)\,dx \leq\liminf_{n\to\infty} \int_\mathbb{R} f_n(x)\,dx \end{align} So: \begin{align} \lim_{n\to\infty} \int_\mathbb{R} f_n(x)\,dx =\int_\mathbb{R}f(x)\,dx \end{align}

- Counterexample (b). Do it yourself. See for example @mathcounterexamples.net

- Proof (c). Since $e^t$ is integrable on $[a,b]$ the result follows from Dominated Convergence Theorem.