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Assume that $k \in \mathbb{N}$ and $f : \mathbb{R}^d \rightarrow [0,\infty)$ is lower semicontinuous, i.e. $f(x) \leq \liminf_{y \rightarrow x} f(y)$ for all $x \in \mathbb{R}^d$.

Does there exist an increasing sequence of $k$-times continuously differentiable functions $(g_n)_n \subset C^k(\mathbb{R}^d ; [0,\infty))$ that converges pointwise to $f$, i.e. $g_n(x) \leq g_{n+1}(x)$ for all $x \in \mathbb{R}^d$ and $n \in \mathbb{N}$ as well as $\lim_{n \rightarrow \infty} g_n(x) = f(x)$ for all $x \in \mathbb{R}^d$?

My intuition would be yes, since we could tile $\mathbb{R}^d$ into dyadic cubes, take for the center of every cube the minimum of the values of $f$ on neighboring cubes and then interpolate between these center points with nice $C^\infty$ functions. Unfortunately this approach sounds very technical to me and I'm wondering whether there is something more elegant.

Michael Greinecker
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x_Y_z
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  • I can construct such an Family, but they are not positive everywhere. Maybe with some modification we can turn then all positive. Anyway the method is something close to waht you have proposed. Do you want me to write it? – Tomás Mar 13 '13 at 13:07
  • Unfortunately, it is important to me that the approximating functions are non-negative. But I am nevertheless interested in your method. Maybe you can sketch it as a comment. – x_Y_z Mar 13 '13 at 14:45
  • Let $F=\{g\in C^\infty:\ g\leq f\}$. Note that $F\neq\emptyset$. Let $B_k$ be an enumeration of all balls with rational center and ratios. Define $g_{i,k}\in F$ in such a way that $g_{i,k}=\inf_{B_k} u-\frac{1}{i}$ in $\frac{1}{2}B_k$. Now define $g_j=\max_{1\leq i,k\leq j} g_{i,j}$. Finally define $h_j=g_j-\frac{1}{j}$, so $h_j$ is the required family, however they don't need to positive everywhere. But if your function is bounded below by an positive constant, then for big $j$, they are positive everywhere. – Tomás Mar 13 '13 at 14:58
  • Would you be kind enough to define $R^d$ for me? – Betty Mock Jan 01 '14 at 08:35
  • @BettyMock It is safe to assume that $\mathbb R^d$, with $\mathbb R$ set in blackboard bold font, means the $d$th Cartesian power of $\mathbb R$; i.e., the set of $d$-tuples $(x_1,\dots,x_d)$ with each $x_i\in\mathbb R$. – Post No Bulls Jan 05 '14 at 19:15
  • @PostNoBulls of course I usually call the $R^n$, so I thought the d might have some special connotation. – Betty Mock Jan 07 '14 at 03:38
  • Look at http://math.arizona.edu/~faris/realb.pdf corollary 1.9. It seems to me to imply that on a compact set the lower semicontinuous functions will be dense in the continuous functions. Pushing this idea seems like it might be fruitful. – Betty Mock Jan 07 '14 at 03:54
  • @Betty Mock of course since any continuous function is lower semicontinuous .... actually other direction is nontrivial and might helpful, are continuous function dense in lsc function, answer is No... for supremum norm – Red shoes Jul 07 '17 at 07:05

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I came across this looking for a wrong theorem. Sorry if it is too late. I would avoid taking sups because they may not preserve smoothness. But if you know an increasing sequence of continuous functions that converge to your lsc function, you may obtain smooth ones by removing $2^{-n}$ to the current function, approximate it within $2^{-n-2}$ by a smooth function (but in the whole space $R^d$ a brutal convolution will not work, I don't know a better way than using partitions of unity before you convolve). Anyway, your new sequence is smooth and still increasing, and converges to the same limit. Agreed, this will not be nonnegative if your initial function $f$ was zero somewhere. For that case I am afraid I see no way to avoid doing this by hand, working on the open set where $f > 2^{-n}$, doing the same sort of thing as above there, and gluing by hand in the remaining region. (Sorry, did not spend too much time).

Cameron Buie
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G. David
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