which of the following statements are true or false?

Question

let $f_n(x)$, for $n\ge 1$, be a sequence of continuous non negative functions on $[0,1]$, such that

$$\ \lim_{n\to \infty} \int_0^1 f_n(x) \, dx=0.$$

which of the following statements is always correct?

A. $f_n\to 0$ uniformly on $[0,1]$

b. $f_n$ may not converge uniformly but converges pointwise.

c. $f_n$ will converge pointwise and limit may be non zero.

d. $f_n$ is not guaranteed to have a pointwise limit.

can anyone help me to solve above problem.

i know that if $f_n$ converges uniformly then limit of interal converges to interal of limit.

Answer 1

HINT:

If $f_n(x)=(1-x)^n$, then

$$\int_0^1 (1-x)^n\,dx =\frac{1}{n+1}\to 0 \,\,\text{as}\,\,n\to \infty$$

But

$$\lim_{n\to \infty }f_n(x)=\begin{cases}1&,x=0\\\\0&,1\ge x>0\end{cases}$$

and therefore, the convergence of $f_n(x)$ is clearly not uniform.

SPOLIER ALERT Scroll over the highlighted area to reveal the solution

If $f_n(x)=\sin(nx)g(x)$,where $g(x)$ is continuously differentiable, then it is easy to show that $$\int_0^1 f_n(x)\,dx\to 0$$(use the Riemann Lebesgue Lemma or integrate by parts). Noting that $\lim f_n(x)$ fails to exist for $x\ne 0$ and we conclude the answer is (d) $f_n$ is not guaranteed to have a pointwise limit.

Answer 2

The answer is (d) in case you still can not see it by considering the function

$$ \sin(nx) . $$

Answer 3

Hint

Take for the sequence of functions $(f_n)$ a hill which is getting narrower, rolling around and around in the interval $(0,1)$ and such that it's integral is equal to a sequence of reals converging to $0$. I let you write a precise case of such a sequence of functions.

$(f_n)$ doesn't converge pointwise proving that all four assumptions are wrong.