Is it true that for each bounded continuous function $f:\mathbb R \to \mathbb R$, we can find a set of analytic functions $g_i:\mathbb R \to \mathbb R, i=1,2,…$ such that $g_i$ uniformly converges to $f$ ?
[Math] Is it true that for each bounded continuous function we can find a set of analytic functions to uniformly converge it
real-analysis
Related Solutions
$C_u(\mathbb R)^*$ is essentially the space of complex measures on $\beta \mathbb Z\coprod (\beta\mathbb Z\times(0,1)).$ Here $\beta \mathbb Z$ is the Stone-Čech compactification of $\mathbb Z,$ and the $\coprod$ denotes disjoint union.
One can identify $C_u(\mathbb R)$ with $C_0(\beta \mathbb Z \coprod (\beta \mathbb Z\times (0,1)))$ in the following way: for $f\in C_u(\mathbb R),$ and write $f=g+h$, where $g(n)=0$ for all $n\in \mathbb Z$ and $h$ is continuous and linear on each interval $[n,n+1].$ We will identify $g$ with a function $\tilde g:\beta \mathbb Z\times [0,1]\to \mathbb C$ in the following way: since $f:\mathbb R\to \mathbb C$ is uniformly continuous, the functions $g|_{[n,n+1]}, n\in \mathbb Z$ form an equicontinuous family, considered as functions $g_n\in C([0,1]).$ By Arzelà-Ascoli, the set $\{g_n:n\in \mathbb Z\}$ is precompact in the uniform topology. By the universal property of $\beta \mathbb Z$, there is a unique continuous function $\varphi: \beta \mathbb Z\to C([0,1])$ such that $\varphi(n)=g_n$ for $n\in \mathbb Z.$ Now the function $\tilde g(x,y):=\varphi(x)(y)$ is a continuous function from $\beta \mathbb Z\times [0,1]$ to $\mathbb C$; the joint continuity is obtained by again applying equicontinuity of the family $\{\varphi(x):x\in \beta\mathbb Z\}.$
We have identified $f$ with a pair $(\tilde g, h),$ where $\tilde g: \beta \mathbb Z\times [0,1]\to \mathbb C$, $\tilde g(x,0)=\tilde g(x,1)=0$ for all $x\in \beta \mathbb Z,$ and $h:\mathbb R\to \mathbb Z$ is determined by the sequence of values $h(n),n\in \mathbb Z.$ It's easy to check that every such pair $(\tilde g, h)$ uniquely determines a function $f\in C_u(\mathbb R)$ by $f(n+x)=\tilde g(n,x)+h(n+x), x\in [0,1), n\in \mathbb Z.$
If we use the norm $|(\tilde g, h)|:=|\tilde g|+|h|$ (these are sup norms), the identification $f\leftrightarrow (\tilde g,h)$ is an identification of $C_u(\mathbb R)$ with $C_0(\beta \mathbb Z \coprod (\beta \mathbb Z\times (0,1))),$ as Banach spaces.
So it appears that finding a non-obvious element of $C_u(\mathbb R)^*$ is more or less equivalent to finding a non-obvious element of $C_b(\mathbb Z)^*,$ as Greg predicted.
If your sequence of functions $r_\alpha$ is uniformly equicontinuous, then this result should hold. That is, there should be one modulus of continuity for all functions in the sequence. Note that the sequence of @i707107 does not satisfy this stronger property. The proof goes along the same lines as the proof that C([0,1]) with supremum norm is a Banach (i.e. complete) space.
Best Answer
Convolve it with narrower and narrower Gauss kernels.