Convergence of sequence of functions

Let $\{f_n\}$ be a sequence of functions deﬁned on $[0,1]$. Suppose that there exists a sequence of numbers $x_n$ belonging to $[0,1]$ such that $f_n(x_n)=1$.

Prove or Disprove the following statements.

• a) True or false: There exists $\{f_n\}$ as above that converges to $0$ pointwise.
• b) True or false: There exists $\{f_n\}$ as above that converges to $0$ uniformly on $[0,1]$.
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Welcome to MSE! You might find a warmer response if you 1) reword the question to remove the imperative (and show that you aren't just a homework-copying robot), and 2) show some work – The Chaz 2.0 Sep 30 '11 at 1:54
Please check the quantifiers. Should these be true for all sets $\{f_n\}$ or for some set $\{f_n\}$. My guess is for some set as $f_n=1$ for all $n$ and $x$ satisfies the condition and makes both false. – Ross Millikan Sep 30 '11 at 2:07
@Ross, the questions make sense if "as above" is taken to mean "such that the conditions in the first paragraph are satisfied". – Henning Makholm Sep 30 '11 at 2:31
@HenningMakholm: Arturo Magidin improved the wording after I put in this comment. – Ross Millikan Sep 30 '11 at 2:33

The first one is true. Let $f_n(x)$ be a function that creates a series of triangles with base $[\frac{1}{2^n}, \frac{1}{2^{n-1}}]$ with $n \in \mathbb{N}.$ Allow each triangle to be of area 1 so that they get taller and skinnier as they get closer and closer to zero. Outside of the triangle they are just zero. This is easier to draw a picture of than describe. This converges pointwise to 0.
I think it'd be more intuitive if you set each triangle's height equal to $1$ (so the $x$-coordinate of the tip is the $x_n$ value) instead of the area. – anon Sep 30 '11 at 4:00
And I think it'd be more intuitive if you set $f_n(1/n)=1$ and $f_n(x)=0$ for other values of $x$. :-D – user1551 Sep 30 '11 at 4:55