Give an example of a sequence of continuous functions $f_n:[0,1]\rightarrow [0,1]$ such that $\lim_{n\rightarrow\infty}m(E_n(\varepsilon)) = 0$ for every $\varepsilon >0$ but $\lim_{n\rightarrow\infty}f_n(x)=0$ for no $x$.

Where $E_n(\varepsilon) := \{x\in [0,1] : |f_n(x)|\geq \varepsilon\}$

My intuition for this is that I need this sequence of functions $\{f_n\}$ to map larger and larger sections of $[0,1]$ closer and closer to zero as $n$ increases. While at the same time never allowing any $x$ to actually drop down to zero in the limit. This seems impossible to me! I mean if most of the points in $[0,1]$ must get arbitrarily close to zero, for large enough $n$, this is the very definition of going to zero in the limit.

Ok so even though it seems impossible, I've still be trying to find a function using my bag of tricks:

  • raising functions to $\frac{1}{n}$
  • speeding up the oscillating velocity of sine
  • looking at sections of the exponential function farther and farther down the negative real axis
  • making functions out of the harmonic series or other infinite series
  • trying to show such a function must exist rather than actually constructing one

Anyways, no luck so far. Maybe just a hint would be good for now and then I could come back if I need more help. Thanks.

  • $\begingroup$ Doesn't this answer your question? $\endgroup$ – leo Oct 31 '12 at 4:17

You are thinking about your sequence of functions in a very static way. And what works is the opposite: think of a bump travelling back and forth and getting narrower as time goes by.

Let $g_{c,d}(t)$ be the function $$ g_{c,d}(t)=\begin{cases}\frac{2(t-c)}{d-c},&\ t\in[c,\frac{d+c}2] \\ \ \\ -\frac{2(t-\frac{d+c}2)}{d-c},\ & t\in(\frac{c+d}2,d]\\ \ \\ 0&\ \mbox{ elsewhere }\end{cases} $$ (i.e. a triangle of height one supported on the interval $[c,d]$). Write $\{q_n\}$ for an enumeration of the rationals of $[0,1]$. Now consider the sequence of functions $$ f_{q_n,\frac1m}(t),\ \ \ n,m\in\mathbb N. $$ Then $m(E_{q_n,1/m}(\varepsilon))\leq1/m$ for all $m$. And if you fix $x\in[0,1]$, then we can find a subsequence $\{q_{n_k}\}$ such that $|x-q_{n_k}|<1/2k$, which implies that $f_{q_{n_k},k}(x)\geq1/2$.

By relabeling the sequence $\{f_{q_n,m}\}$ as $f_n$ we get the desired sequence.


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