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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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    @HenningMakholm: Arturo Magidin improved the wording after I put in this comment.2011-09-30

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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.

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    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 $x$. :-D2011-09-30