Sequence of smooth functions converging to sgn(x) - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T00:24:42Z http://mathoverflow.net/feeds/question/88752 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/88752/sequence-of-smooth-functions-converging-to-sgnx Sequence of smooth functions converging to sgn(x) Alex Flint 2012-02-17T19:04:46Z 2012-02-19T19:41:45Z <p>I'm looking for a sequence of smooth functions $f_i(x)$ converging to <a href="http://en.wikipedia.org/wiki/Sign_function" rel="nofollow">Sign</a>$(x)$, each of which additionally have the following property:</p> <p>$$f_i(x_1+x_2) = g_i(x_1, f_i(x_2))$$</p> <p>for some $g_i$</p> <p>Also, is there a name for a function that satisfies just the decomposability constraint above?</p> http://mathoverflow.net/questions/88752/sequence-of-smooth-functions-converging-to-sgnx/88756#88756 Answer by Barry Cipra for Sequence of smooth functions converging to sgn(x) Barry Cipra 2012-02-17T19:50:31Z 2012-02-19T19:41:45Z <p>Let $f_i$ be any sequence of strictly increasing smooth functions that converge to the Sign function, such as $f_i(x) = \tanh(ix)$, and let $g_i$ be defined by $g_i(x,y) = f_i(x+f_i^{-1}(y))$ for $y$s in the range of $f_i$ (e.g., $-1 &lt; y &lt; 1$) and however you like elsewhere (since the problem posits no smoothness or even any continuity conditions on $g$). This takes care of the specific problem.</p> <p>In general, any continuous function satisfying the "decomposability constraint" is either constant or strictly monotonic. Indeed, suppose $f$ is not strictly monotonic, so that (courtesy of the Intermediate Value Theorem) $f(a)=f(b)$ for some $a \ne b$. Then</p> <p>$$f(x+a)=g(x,f(a))=g(x,f(b))=f(x+b),$$</p> <p>from which it follows (by substituting $x-a$ for $x$ in the displayed equation) that $f$ is periodic with period $b-a$. Since $f$ is continuous, there are values $c$ and $d$ between $a$ and $b$ with $f(c)=f(d)$ and $|c-d|$ arbitrarily small (think points to either side of where $f$ takes its maximum value). Hence $f$ is periodic with arbitrarily small period -- which is to say, constant.</p>