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# Infinite ABS Inequality

Prove that $\sum_{k=-N}^{N}|x+k|-|k| \ge x^2$ as $$N\to \infty$$ and find equality cases.

Details and Assumptions

We take the limit as $$N$$ approaches $$\infty$$ because $$\displaystyle\sum_{k=-\infty}^{\infty}|x+k|-|k|$$ is not well-defined.

Note by Daniel Liu
2 years, 5 months ago

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Set : $f_n(x)= \sum_{k=-n}^n |x+k|-|k| =|x|+\sum_{k=1}^n |x+k|+|x-k| -2|k|$ Note that $$f_n(-x)=f_n(x)$$ for any real $$x$$ and any integer $$n>0$$ then we only need to consider the case $$x\geq 0$$, and if $$k\geq x>0$$ then $$|x+k|+|x-k|-2|k|=0$$, therefore for $$n>x$$ : $f_n(x)=x+\sum_{k=1}^{\lfloor x\rfloor}(x+k)+(x-k)-2 k= x+2x \lfloor x \rfloor -\lfloor x\rfloor (\lfloor x\rfloor +1).$

Clearly now, $$f_n(x)= x-\lfloor x\rfloor +2x \lfloor x\rfloor-\lfloor x\rfloor^2$$, the rest should follow easily.

$f_n(x)-x^2 =x- \lfloor x\rfloor + x(\lfloor x\rfloor-x)+\lfloor x\rfloor(x-\lfloor x\rfloor)=(x-\lfloor x\rfloor)(1+\lfloor x\rfloor -x)\geq 0$

Obviously, the equality is when $$x\in \mathbb{Z}$$.

Remark : you don't need $$n\to \infty$$ you just need $$n\geq x$$, and the latter series does make sense and it well-defined (though I'd use the term 'convergent' because well defined usually used for mono-valued functions). · 2 years, 5 months ago

How is the latter sequence defined? I seem to be getting different values for plugging in the same $$x$$ value if I compute it differently. · 2 years, 5 months ago