I want to prove that $\sqrt x $ is sublinear, but i don't know how to do it, precisely i'm having difficulties stating that $\sqrt x $ is sublinear in $[0, 1]$ $$\\ f(x)\ is\ sublinear\ \iff\ \exists\ a,b \in\ \mathbb R\ \mid\ |f(x)| \le a|x| + b $$
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Is this what you are looking for? https://math.stackexchange.com/q/408177/42969 – Martin R Apr 06 '22 at 13:10
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2Does this answer your question? How to prove $\sqrt{x + y} \le \sqrt{x} + \sqrt{y}$? – K.defaoite Apr 06 '22 at 13:22
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No, i added the definition – Iruo Apr 06 '22 at 13:30
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2Forgive me if I'm mistaken, but according to your definition, can't you just let a=100000, and b=1 to create a line that is always above $|{\sqrt{x}}|$ in your given range? – yolo Apr 06 '22 at 13:34
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1This is not a proof, even with $a=10^{100}$. @yolo – nicomezi Apr 06 '22 at 13:44
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@nicomezi I didn't bother with rigour to that excess as I assumed that it was a misnomer on the OP's wording, since it seemed too obvious, – yolo Apr 06 '22 at 13:53
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Your $x$ is non-negative. We need $a$ and $b$ such that
$$ \forall x \ge 0 \quad 0\le\sqrt x \le ax +b.$$
Raise to the power $2$: $$0 \le a^2 x^2 + (2 a b - 1) x + b^2$$
Take for simplicity $a=1$: $$0 \le x^2 + (2 b - 1) x + b^2,$$ and that for $x\ge0$. It is sufficient to take $2b-1\ge0$.
TZakrevskiy
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But then if i use a b>0 the negative x straight line will not be under the graph of $\sqrt x$ like here: https://www.google.com/search?q=square+root+of+x+is+not+lipschitz&rlz=1C1ONGR_itIT974IT974&sxsrf=APq-WBuvlxWBAtnvWD2JYeO7FGYuRBXHTQ:1649253422991&source=lnms&tbm=isch&sa=X&ved=2ahUKEwjT16_jy__2AhWui_0HHSmWA8cQ_AUoAXoECAEQAw&biw=1242&bih=568&dpr=1.1#imgrc=gStxtpyljJLhqM – Iruo Apr 06 '22 at 13:58
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$\sqrt x$ is not defined for $x<0$. The first graph is $\sqrt{|x|}$ @GiuseppeFontanella – nicomezi Apr 06 '22 at 13:59
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i'm sorry i intended that to be sublinear the function has to have its own graph contained between two straight line with opposite angular coefficients like in the link i posted in the previous comment – Iruo Apr 06 '22 at 14:03
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@Iruo Perhaps you should use this condition:
$\forall x_1,x_2, \exists K \mid \left | f(x_{1})-f(x_{2}) \right |\leq K\left | x_{1}-x_{2} \right |$
– yolo Apr 06 '22 at 14:10 -
@Iruo then you need to ask for Lipschitz continuity, not sublinearity; square root is not Lipschitz continuous on $[0,\infty)$. – TZakrevskiy Apr 08 '22 at 08:30