Let $f_n:[1,2]\to[0,1]$ be given by $f_n(x)=(2-x)^n$ for all non-negative integers $n$. Let $f(x)=\lim_{n\to \infty}f_n(x)$ for $1\le x\le 2$. Then which of the following is true?
(a) $f$ is continuous on [1,2]
(b) $\lim_{n\to \infty}\int_1^2f_n(x)dx=\int_1^2f(x)dx$
(c)$f_n$ converges uniformly to $f$ on $[1,2]$ as $n \to \infty$
(d) for any $a\in(1,2)$ we have $\lim_{n \to \infty}f_n'(a)\neq f'(a)$
Solution:
$f_n(x)$ converges to $f(x)=1$ when $x=1$ and $f(x)=0$ when $x\in(1,2]$. Hence option (a), (c) are clearly incorrect.
Now for any $a\in(1,2)$ $$\lim_{n\to \infty}f_n'(a)=\lim_{n \to \infty}-n(2-a)^{n-1}=0$$ and $$f'(a)=0 \;\;\forall \;\;a\in(1,2) $$ which implies $\lim_{n \to \infty}f_n'(a)= f'(a)$. Hence option (d) is incorrect as well.
Now, for part (c):
$$lim_{n\to \infty}\int_1^2f_n(x)=\lim_{n \to \infty}\frac{1}{n+1}=0$$ and $$\int_1^2f(x)=\int_1^20dx=0$$
Hence option (c) is correct.
Have I done everything correctly here? Also, i think $\lim_{n\to \infty}\int_1^2f_n(x)dx=\int_1^2f(x)dx$ happens only when $f_n(x)$ converges uniformly to $f(x)$. But here it doesn't converge uniformly then also this result holds, is there any explanation for that? Please help me in this question.
Best Answer
What you have done is correct.
b) is true. Just evaluate the integral explicitly. $\int_1^{2} f_n(x) dx=\int_1^{2} (2-x)^{n} dx=\int_{0}^{1} y^{n} dy =\frac 1{n+1 } \to 0$. And $\int_1^{2} f(x) dx=0$.
Uniform convergence is a sufficient condition for the convergence of the integrals but it is not a necessary condition.