Let $f$ be a measurable function on $[0,1]$. Is there a sequence infinitely differentiable $f_n$ such that one of

  • $f_n\rightarrow f$ pointwise
  • $f_n\rightarrow f$ uniformly
  • $\int_0^1|f_n-f|\rightarrow 0$

is true?

  • 1
    $\begingroup$ Is $f_n$ supposed to be a sequence of $C^\infty$ functions? $\endgroup$ – spitespike May 1 '13 at 3:31
  • $\begingroup$ @spitespike Yes. Sorry for the ambiguity. I have edited the question. $\endgroup$ – Spook May 1 '13 at 3:54
  • $\begingroup$ Have you had any ideas about this? Can you pin down the problems you're having at all? $\endgroup$ – Kevin Carlson May 1 '13 at 4:09
  • $\begingroup$ @Montez I am not sure about general conclusions but the first and second holds if $f$ is continuous in some open set. The third holds if $ f \in L^1 $. In both cases to get $f_n$ you have to convolute $f$ with a mollifier supported in $ B(0,1/n)$. $\endgroup$ – smiley06 May 1 '13 at 5:01
  • $\begingroup$ Is $f$ also supposed to be integrable, in the sense that $\int |f| < \infty$? $\endgroup$ – Jakub Konieczny May 1 '13 at 6:35

Uniform convergence is surely too much to ask for. As Wikipedia suggests, uniform convergence theorem assures that the uniform limit of continuous functions is again continuous. Hence, as soon as $f$ is discontinuous, all hope of finding smooth $f_n$ uniformly convergent to $f$ is gone.

The statement involving the integral is true (if we additionally assume $\int |f| < \infty$, at least), and follows from a more general fact that the continuous functions are dense in $L^1([0,1])$ (integrable functions with norm given by $||f|| = \int |f|$). A possible way to check this is the following. First, measurable functions can be arbitrarily well approximated by simple functions (the ones of the form $\sum a_i \chi_{A_i}$, with $A_i$ - measurable sets). Thus, if we are able to approximate the function $\chi_A$ arbitrarily well by continuous functions, then we are done. For this, notice that $A$ can be approximated by an open set $U$: for any $\varepsilon > 0$, there is open $U$ with $\lambda(A \triangle U) < \varepsilon$. Now, $U$ is open, so you can express it as a sum of intervals: $U = \bigcup I_n$, $I_n$ - open interval, disjoint from the $I_m$, $m\neq n$. Now, $\chi_I$ can be approximated by smooth functions by using classical bump functions. A lot of details would have to be filled in, but it should be clear that a measurable function can indeed be arbitrarily well approximated by smooth ones, in $L^1$.

For pointwise convergence, I think you can use a reasoning as just offered for $L^1$. I also believe you can use mollifiers. Since you only asked if one of the statements can be made true, I shall not go into more detail. Also, I am not quite sure what background to assume, and I am more that sure there are a lot of other users who have much better understanding of these issues.


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