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Prove $x+\sin\sqrt{x}$ is uniformly continuous on $[0,\infty)$. Here's is my proof, but I'm worried of me $\delta$ choice being incorrect.

If $x_1,x_2>0$ then $$|f(x_1)-f(x_2)|=|x_1+\sin\sqrt{x_1}-x_2-\sin\sqrt{x_2}|$$

Since $-1\le\sin x\le1$:


Then let $\delta=\epsilon-2$ so that-

$\forall x_1,x_2 |x_1-x_2|<\delta => |f(x_1)-f(x_2)|\le|x_1-x_2|+2<\epsilon-2+2=\epsilon$

End of proof.

My problem is that my $delta$ is $\delta=\epsilon-2$, but $\delta$ is positive, so there is no $\delta$ for $\epsilon=1$ because then $\delta$ will be negative. So is my solution wrong and if so what the error?


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Why not? It does in the question. – Harold Jan 4 '13 at 2:39
It's wrong precisely because your $\delta$ might be negative. – David Mitra Jan 4 '13 at 2:44
@DavidMitra Ok, so I thought. I tried solving it again and had no idea how to do it... – Harold Jan 4 '13 at 2:45
Do you know that the sum of two uniformly continuous functions is uniformly continuous? And that the composition of uniformly continuous functions is uniformly continuous? If you can't use these facts, then this post may give you some ideas. – David Mitra Jan 4 '13 at 2:50
@DavidMitra Good idea, but you do realize who asked the question in that other other post, yes? – Ellie Kesselman Jan 4 '13 at 2:56

Yes, $\delta$ has to be positive here; your solution is wrong precisely because yours isn't.

If you can't use the facts mentioned in my comment above, the following outline may be used to show your result:

$\ \ \ $1) Show that the functions $h(x)=\sqrt x$ and $g(x)=\sin x$ are uniformly continuous on $[0,\infty)$.

$\ \ \ $2) Let $\epsilon>0$.

$\ \ \ $3) Choose a $0<\delta_1$ so that $|\sin x_1-\sin x_2|<\epsilon/2$ whenever $|x_1-x_2|<\delta_1$.

$\ \ \ $4) Choose $0<\delta_2$ so that $|\sqrt x_1-\sqrt x_2|<\delta_1$, whenever $|x_1-x_2|<\delta_2$.

$\ \ \ $5) Set $\delta=\min\{\epsilon/2,\delta_2\}$.

$\ \ \ $6) Suppose $|x_1-x_2|<\delta$ and consider the inequality $$|f(x_1)-f(x_2)|\le |x_1- x_2|+| \sin \sqrt x_1-\sin \sqrt x_2|.$$

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