Approximating measurable functions on $[0,1]$ by smooth functions.

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?

• Is $f_n$ supposed to be a sequence of $C^\infty$ functions? – spitespike May 1 '13 at 3:31
• @spitespike Yes. Sorry for the ambiguity. I have edited the question. – Spook May 1 '13 at 3:54
• Have you had any ideas about this? Can you pin down the problems you're having at all? – Kevin Carlson May 1 '13 at 4:09
• @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 $B(0,1/n)$. – smiley06 May 1 '13 at 5:01
• Is $f$ also supposed to be integrable, in the sense that $\int |f| < \infty$? – Jakub Konieczny May 1 '13 at 6:35

1 Answer

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.