Certain uniform approximations of a bounded uniformly continuous function

Question

Let $u:\mathbb{R}^N \to \mathbb{R}$ be a bounded and uniformly continuous function.

(a) Do there exist and (b) how can we construct three families of functions $u_\epsilon:\mathbb{R}^N \to \mathbb{R}$ ($\epsilon \to 0^+$) such that $u_\epsilon \to u$ uniformly (not just locally uniformly) and respectively

1. $u_\epsilon \in C^\infty(\mathbb{R}^N)$
2. $u_\epsilon \in \mathrm{Lipschitz \ and \ bounded}(\mathbb{R}^N)$
3. $u_\epsilon \in C^2(\mathbb{R}^N)$ bounded and with bounded derivatives up to the second?

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I'm aware that if $u$ is continuous then the family of the standard mollifications tends to $u$ locally uniformly.

Answer 1

1. The standard proof of the convergence of convolutions works here if you choose a compactly supported mollifier $\rho$. For every $\delta>0$ take the $\epsilon$ given by the uniform continuity of $u$, i.e. $|x-y|\leq \epsilon\implies |u(x)-u(y)|\leq \delta$. Choose $\rho_\epsilon$ mollifier such that $\text{supp}\,\rho_\epsilon\subset B(0,\epsilon)$. Then \begin{align} |u*\rho_\epsilon(x)-f(x)|&=\left|\,\int\limits_{\mathbb R^n} (u(x-y)-u(x))\rho_\epsilon(y)dy\right|\\ &\leq \int\limits_{\mathbb R^n}|u(x-y)-u(x)|\rho_\epsilon(y)dy\\ &=\int\limits_{B(0,\epsilon)}|u(x-y)-u(x)|\rho_\epsilon(y)dy\leq\delta. \end{align}

2. To obtain $u_\epsilon$ evaluate $u$ at the points of $L_\epsilon:=\epsilon\mathbb Z^n$, divide every $\epsilon$-cube in simplices and interpolate affinely inside every simplex. Uniform continuity gives Lipschitzianity of $u|_{L_\epsilon}$, while the Lipschitzianity of the interpolation follows by the fact that $$\|A\|_*:=\sum_{i=1}^n|Av_i|$$ is a norm on the $n\times n$ matrices if $v_1,\ldots,v_n$ are fixed linearly independent vectors, and in finite dimension all norms are equivalent. Apply this with $v_i$ equal to the edges of the simplices.

   EDIT: adding some details. Anyway, now that I won't use these functions for point 3 anymore, I think the cleanest way is the one suggested by s.harp in the comments.

   By u.c. we have that $\exists\epsilon$ such that $|u(x)-u(y)|\leq 1$ if $|x-y|\leq \epsilon$. This implies that $u|_{L_\epsilon}$ is $1/\epsilon$-Lipschitz. Now divide each cube of $L_\epsilon$ in a finite number of simplices ("hypertetrahedra") whose vertices lie in $L_\epsilon$, and interpolate linearly inside each of them (in dimension $2$ this would mean divide every square in two right triangles). Now use this:

   Lemma Let $S=\{0,v_1,\ldots,v_n\}\subset \mathbb R^n$ with $v_1,\ldots,v_n$ unit vectors linearly independent, and suppose $u\colon S\to\mathbb R$ is $L$-Lipschitz on $S$, i.e. $$L=\text{Lip}(u,S):=\sup_{x,y\in S,x\neq y}\frac{|f(x)-f(y)|}{|x-y|}.$$ Then the unique affine extension of $u$ to $\text{conv}(0,e_1\ldots,e_n)$ is $CL$-Lipschitz, where $C$ depends only on the vectors $v_1,\ldots,v_n$ and is invariant under isometries and dilations.

   Proof. Call $A$ the linear interpolation. The Lipschitz constant of $A$ coincides with the operatorial norm $\sup_{|x|=1}|Ax|$, which is equivalent to $\|A\|_*$ defined above because of the finitness of the dimension, which in turn is less than $n\text{Lip}(u,S)$.

   From this you obtain that the Lipschitz constant of the extension is controlled by the Lipschitz constant of $f|_{L_\epsilon}$, which is finite.

3. This is not possible in general if you require equiboundedness of second derivatives. Consider in $\mathbb R$ the function $u(x)=\sqrt{|x|}$. Suppose that $u_\epsilon\to u$ uniformly and that $u_\epsilon$ are $C^2$. Then, by the uniform convergence, $u_\epsilon'(x)$ will eventually (in $\epsilon$) be really high near positive zero, and really low near negative zero (apply Lagrange). In particular for every constant $M$ we can find points $x,y$ such that $|x-y|\leq 1$ and such that for every sufficiently small $\epsilon$ we have $|u'_\epsilon (x)-u_\epsilon'(y)| \geq M$. However, by the FTC $$|u'(y)-u'(x)|=\left|\int_x^y u''(t)dt\right|\leq \|u''\|_\infty|x-y|$$ which is a contradiction.

   If instead you just require boundedness of each second derivative, I believe that a suitable convolution with the functions $u_\epsilon$ at point 2 will do. This essentially because the change in the first derivatives of $u_\epsilon$ between two near simplices is at most given by twice the Lipschitz constant, and happens at lengthscale $\epsilon$. Therefore a convolution with a mollifier with support of lengthscale $\sim\epsilon$ will make a smooth transition of first derivatives between a simplex and a neighbour, hopefully mantaining a second derivative of order $\sim 1/\epsilon$.

   EDIT Without the uniform boundedness it is sufficient to use the same convolutions as point 1. Indeed $$\|\partial^2_{ij}(u*\rho)\|_\infty=\|u*(\partial^2_{ij}\rho)\|_\infty\leq \|u\|_\infty \|\partial_{ij}^2\rho\|_1$$ which is finite if $\rho$ is regular and has compact support.