Let $$ f_n(x) := \begin{cases} 1 &\text{for $x$ in } \left(0, \frac{1}{n}\right)\\ 0 &\text{$x$ elsewhere in } [0,1] \end{cases}. $$ Show that $\{f_n\}_{n=1}^\infty$ is a decreasing sequence of discontinuous functions that converges to a continuous limit function, but the convergence is not uniform on $[0,1]$.

I don't really know what do for this question. It's not actually homework, I'm just trying to do practice questions before my final. Any help is appreciated.

  • It is decreasing because there are only two values, and $f_n^{-1}(1)$ is decrasing sequence of sets.
  • It converges pointwise to $0$ on $(0,1)$ because for a fixed $x$, for $n$ big enough you always has $x\notin (0,1/n)$
  • $\sup_{x\in (0,1)} |f_n(x)| = 1 \nrightarrow 0$, hence the convergence is not uniform.
  • $\begingroup$ can you elaborate on why it's decreasing? $\endgroup$ – User38 Apr 21 '14 at 22:58
  • $\begingroup$ $f^{-1}(\{1\}) = (0, 1/n)$. this means that the function gets the highest value on this interval, which gets smaller and smaller $\endgroup$ – mookid Apr 21 '14 at 23:48

Well, $\sup\limits_{x\in[0,1]}|f(x)-f_n(x)|$ is always equal to...? Hence it doesn't go to zero, and convergence is not uniform.


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