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(http://math.stanford.edu/~ksound/Math171S10/Hw8Sol_171.pdf)

Prove for all $e > 0,$ there exists $d > 0$ : for all $x, y \ge 0$, $|x - y| < d \implies |\sqrt{x} - \sqrt{y}| < e$.

(a) Given $\epsilon>0$, pick $\delta=\epsilon^{2}$. First note that $|\sqrt{x}-\sqrt{y}|\leq|\sqrt{x}+\sqrt{y}|$.
Hence if $|x-y|<\delta=\epsilon^{2}$, then $ |\sqrt{x}-\sqrt{y}|^{2}\leq|\sqrt{x}-\sqrt{y}||\sqrt{x}+\sqrt{y}|=|x-y|<\epsilon^{2}. $
Hence $|\sqrt{x}-\sqrt{y}|<\epsilon$.

1. Where does $|\sqrt{x}-\sqrt{y}|\leq|\sqrt{x}+\sqrt{y}|$ issue from? How to presage this presciently?
Yes...can prove it. Square both sides. By dint of $|a|^2 = (a)^2$: $|\sqrt{x}-\sqrt{y}|^2 \leq|\sqrt{x}+\sqrt{y}|^2 \iff (\sqrt{x}-\sqrt{y})^2 \leq(\sqrt{x}+\sqrt{y})^2 \iff - 2\sqrt{xy} \le 2\sqrt{xy} \\ \iff 0 \le 4\sqrt{xy}. █ $.

2. Figue or Intuition please for $|\sqrt{x}-\sqrt{y}|\leq|\sqrt{x}+\sqrt{y}|$? Feels fey.
I know $|x + y| \le |x| + |y| \iff |x - y| \le |x| + |y|$.

3. For scratch work, I started with $|\sqrt{x} - \sqrt{y}| < e$. But answer shows you need to start with $|\sqrt{x} - \sqrt{y}|^2$. How to presage this vaticly? What's the scratch work for finding $d = e^2$?

4. How does $d = e^2$ prove uniform continuity? I know this proves uniform continuity.
Solution needs to prove $d = e^2$ doesn't depend on x or y?

3 Answers3

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When $0\leq x\leq y$ then

$$\left(\sqrt{\mathstrut y}-\sqrt{\mathstrut x}\right)^2=y-2\sqrt{\mathstrut xy}+x\leq y-x\ .$$ It follows that $$\sqrt{\mathstrut y}-\sqrt{\mathstrut x}\leq\sqrt{\mathstrut y- x}\ .$$ This implies $$\left|\sqrt{\mathstrut y}-\sqrt{\mathstrut x}\right|\ \leq\ \sqrt{|y-x|}$$ for arbitrary $x$, $y\in{\mathbb R}_{\geq0}$ and therewith the uniform continuity of $\sqrt{\cdot}\>$.

Christian Blatter
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Or just prove using the binomial theorem that for $0\le x\le y=x+h$, you get

$$\sqrt[n]{x+h}\le\sqrt[n]{x}+\sqrt[n]h,$$

so that finally for any $x,y\ge 0$

$$\left|\sqrt[n]y-\sqrt[n]x\right|\le\sqrt[n]{|y-x|}.$$

Now compare with the definition of Hölder continuity with Hölder index $α\in(0,1]$,

$$|f(y)-f(x)|\le C\cdot|y-x|^\alpha$$

and use that Hölder continuity implies uniform continuity.

@1) already answered sufficiently in question.

@2) The use of the binomial theorem for an arbitrary power $n$, essentially $A^n+B^n\le(A+B)^n$, with $A=\sqrt[n]x$ and $B=\sqrt[n]{y-x}$ generalizes the method given in the question.

@3) In this generalized form as equation for Hölder continuity, the motivation is much clearer.

@4) choosing $δ=ϵ^n$ results in the conclusion that if $|x-y|<δ$ for any $x,y\ge 0$, then $\left|\sqrt[n]y-\sqrt[n]x\right|<\sqrt[n]{ϵ^n}=ϵ$, and this proves uniform continuity.

Lutz Lehmann
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    How does this actually answer the questions asked by the OP? – Willie Wong Feb 19 '14 at 09:49
  • Because it shows uniform Hölder continuity with Hölder index $1/n$, i.e., even something stronger than just uniform continuity. – Lutz Lehmann Feb 19 '14 at 10:07
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    The OP already wrote down (a sketch of) a proof, which is essentially the same as what you gave above. The OP asks four very specific questions about that proof, none of which is addressed in your supposed answer. Let me repeat: exactly how does this answer the questions that the OP asked? – Willie Wong Feb 19 '14 at 13:03
  • The sketch of the proof in the question is very much bound to the square root. Clever tricks that work in this case fail in higher order cases. The generalized approach is IMO much clearer. – Lutz Lehmann Feb 19 '14 at 13:16
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2. You know that $\sqrt{x}$ and $\sqrt{y}$ are non negative. The sum of two non-negative numbers is always at least as great as their difference. Alternatively, $|x| + |y| = |x + y|$ for non negative $x$ and $y$. Thus $|\sqrt{x} - \sqrt{y}| \le |\sqrt{x}| + |\sqrt{y}| = |\sqrt{x} + \sqrt{y}|$.

3. Practise! The solution is not obvious.

4. $\delta$ doesn't depend on $x$ or $y$. It is a function solely of $\epsilon$. Thus, the same $\delta$ works for any $x$ and $y$, which is what is required for uniform continuity.

user246852
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