Sequence of real analytic functions $f_{n}(x):\mathbb{R} \rightarrow \mathbb{R}$ converges uniformly to absolute value function $f(x)=|x|$

Question

Construct a sequence of functions $f_{n}:\mathbb{R} \rightarrow \mathbb{R}$ such that each $f_{n}$ is real analytic, and moreover $f_{n}(x)$ converges uniformly to $f(x)=|x|$ on $\mathbb{R}.$

Answer 1

Note that the uniform limit of these real analytic functions is not analytic.This never happens with complex analytic functions. we have $|x|=\ x \ sgn(x)$ where $\ sgn(x)$ is the sign function

The sequence $(x \tanh(n x))_n$ converges uniformly to $\ sgn(x)$ on every >bounded interval.If you want it uniform on all of $\ \mathbb{R}$ multiply by $\ (1+x^2)^{-1}$.see 1.

I don't know ho to prove the previous fact. Also in an answer to another question posted here see 2.

it says that if we have equicontinuity then we can apply Arzelà–Ascoli theorem we can find a homeomorphism $\phi : [0, 1] \to \mathbb{R}^{\ast}$ such that $f_n \circ \phi$ s equicontinuous. Then we >can apply AA to $(f_n \circ \phi)$ to obtain a uniformly convergent subsequence $(f_n \circ \phi)$. Now it is clear that $f_n \circ \phi$ itself is also uniformly convergent.see 2.

1. @MISC {1121184,
TITLE = {A uniformly convergent sequence of real analytic functions which does not converge to a real analytic function},
AUTHOR = {quid (http://math.stackexchange.com/users/85306/quid)},
HOWPUBLISHED = {Mathematics Stack Exchange},
NOTE = {URL:http://math.stackexchange.com/q/1121184 (version: 2015-01-27)},
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2. @MISC {158098,
TITLE = {Subsequence Convergence},
AUTHOR = {Sangchul Lee (http://math.stackexchange.com/users/9340/sangchul-lee)},
HOWPUBLISHED = {Mathematics Stack Exchange},
NOTE = {URL:http://math.stackexchange.com/q/158098 (version: 2012-06-14)},
EPRINT = {http://math.stackexchange.com/q/158098},
URL = {http://math.stackexchange.com/q/158098}} 

Answer 2

Try:

$f_n(x)=\sqrt {x^2+\frac{1}{n}}$