Is $(f_n)$ pointwise convergent?

Question

Let $f_n(x)$, for all n>=1, be a sequence of non-negative continuous functions on [0,1] such that

$$\lim_{n→\infty}\int^1_0 f_n (x)dx=0$$

Which of the following is always correct ?

A. $f_n→0$ uniformly on $[0,1]$

B. $f_n$ may not converge uniformly but converges to $0$ pointwise

C. $f_n$ will converge point-wise and limit may be non zero

D. $f_n$ is not guaranteed to have pointwise limit.

I can find example where all four statements are true but can't find any counterexample.

Option A is certainly false. Consider a function (geometrically) a isosceles triangle whose one vertex is on $0$, other on $2/n$. And height $1$ unit. Its area is $2/n →0$ . But $f$ is not uniformly continuous

For others, i am completely stuck. I need ideas to solve this question rather than solution.please help!.

Answer 1

Expanding on the comment I made.

Hint: Think of placing the peak of your isosceles triangle alternating between two (or maybe infinitely many) pre-determined points.

Solution: A more elaborate construction of isosceles triangles suffice to build a sequence of $f_n$ that does not converge. Each $f_n$ can be for instance: non-zero only on an interval $I_n = (a_n - \delta_n, a_n + \delta_n)$, piecewise linear, with $f_n(x)$ being $0$ on the extremes of the interval and $f_n(a_n) = 1$. Notice that we have freedom to choose $a_n$ and $\delta_n$. Since $\int f_n = \delta_n$, we just need to ask $\delta_n \to 0$ for the limit above to hold.

The main point to force that $f_n$ do not converge pointwise is the choice of $a_n$. If for instance $a_{2n} = \frac13$ and $a_{2n+1} = \frac23$, then $f_n(\frac13)$ and $f_n(\frac23)$ will oscilate between $0$ and $1$ infinitely many times, while for all other $x$ we have $f_n(x) = 0$.

To Think: If you want to build a sequence of functions with infinitely many points that do not converge, you can try a different choice of $a_n$. For instance, let $q_i$ be an enumeration of $\mathbb{Q} \cap (0, 1)$ and $p_i$ be the $i$-th prime number. Then, we can try defining $a_n = q_i$ iff $p_i$ is the smallest prime that divides $n$.

Answer 2

A. You could also consider $f_n(x) = x^n.$

B. Let $f_n$ be the line-segment graph connecting the points $(0,1), (1/n,0), (1,0).$ This sequence does not converge to $0$ at $x=0.$

C. Let $f_n$ be as in B. Then consider the sequence $f_1,0,f_2,0,f_3,0,\dots.$ At $x=0,$ this sequence is $1,0,1,0,\dots.$

D. See C.

Answer 3

I can find example where all four statements are true ...

All four?

It seems that the statements are worded in a way that (independent of the rest of the question) exactly one of the four statements must be true.

... but can't find any counterexample.

I even found an example for $f_n$ that does not converge in any point $x$:

Let $f^*_{c,w}(x)$ with $x\in [-10,10]$ be defined as the function with the graph connecting the points:

$(-10,0)$, $((\frac c{2w}-\frac 1w),0)$, $((\frac c{2w}),1)$, $((\frac c{2w}+\frac 1w),0)$, $(10,0)$.

Use $f_{c,w}(x)=f^*_{c,w}(x)$ for $x\in[0,1]$.

Now use the following sequence:

$f_{0,1}$, $f_{1,1}$, $f_{2,1}$,
$f_{0,2}$, $f_{1,2}$, $f_{2,2}$, $f_{3,2}$, $f_{4,2}$,
$f_{0,4}$, $f_{1,4}$, ..., $f_{8,4}$,
$f_{0,8}$, $f_{2,8}$, ..., $f_{16,8}$,
...
$f_{0,(2^k)}$, $f_{1,(2^k)}$, ..., $f_{(2^{k+1}-1),(2^k)}$, $f_{(2^{k+1}),(2^k)}$,
...

If I'm correct this example can be modified by using $f_{c,w}(x)=\sqrt wf^*_{c,w}(x)$. The resulting sequence should meet the conditions and for every value of $x\in[0,1]$ the sequence $f_1(x)$, $f_2(x)$, $f_3(x)$ ... should even be unbounded!

(However maybe I made a mistake here.)