/sci/ - Science & Math

>>12421155
Assume [math]g(0) = a > 0[/math] (the larger than zero part is for convenience and not actually a vital part of the proof. Considering the other two cases by immitating the proof below is your homework.). Then there is some [math]\epsilon[/math] such that [math]|x| < \epsilon[/math] implies [math]g(x) > a/2[/math]. There is also an [math]\epsilon ' > \epsilon[/math] such that [math]|x| < \epsilon ' [/math] implies [math]g(x) > a/4[/math].
Now we consider the function [math]f(x) = 0[/math] if [math]x < \epsilon[/math] or [math]x > \epsilon '[/math], [math]f(x) = x - \epsilon[/math] if [math]\dfrac{\epsilon' + \epsilon}{2} \geq x \geq \epsilon[/math] and [math]f(x) = \epsilon ' - x[/math] if [math]\epsilon ' \geq x > \dfrac{\epsilon' + \epsilon}{2}[/math]. Basically, it's a ramp that starts going up at [math]\epsilon[/math], peaks at [math]\dfrac{\epsilon' + \epsilon}{2}[/math] and returns to zero at [math]\epsilon'[/math].
Trivially, [math]f(0) = 0[/math], but [math]\langle f, g \rangle > 0[/math], a contradiction.