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My attempt:

Given any $x\in [1,\infty)$, let $y = x+k$ for any $k>0$. Then

$|f(y)-f(x)| = \sqrt{(x+k)^3}-\sqrt{x^3} \geq\sqrt{(1+k)^3}-\sqrt{1}$

because $x\geq 1$ and $\sqrt{x^3}$ is a positive, increasing function on $[0,\infty)$.

We then have that

$\sqrt{(1+k)^3}-\sqrt{1} \geq \sqrt{1^3}+\sqrt{k^3}-\sqrt{1}=\sqrt{k^3}$

Hence, for any $|x-y|=k$, we have $|f(x)-f(y)|\geq\sqrt{k^3}$, showing this cannot be uniformly continuous.

I don't feel particularly confident about the last line especially. Does this make sense?

Any help appreciated!

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    "We then have that the above inequality is $\le$ ..." This line is unclear. Which expression are you saying is $\le \sqrt {1^3} + \sqrt{k^3} - \sqrt 1$? – 2017-02-01
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    @tilper Fixed the typo, thank you! – 2017-02-01
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    suggestion: what about using the mean value theorem instead? – 2017-02-01
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    $\sqrt{x}$ is a positive and increasing function on $[0,\infty)$ as well, but it holds $\sqrt{x+k} - \sqrt{x} < \sqrt{1+k} - \sqrt{1}$ for $x > 1$. What you need is some argument that the derivative of your function is still increasing, which can then be used with the mean value theorem as suggested. – 2017-02-01
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    You want to show that there exists an $\epsilon$, (say, $1$), such that there is no $\delta>0$ such that $|x-y| < \delta$ necessarily implies $|f(x) - f(y)| < \epsilon$. I don't think that your final inequality implies that - right? After all, $k$ can be small. My suggestion by the way was to use the MVT for the initial problem. – 2017-02-01

2 Answers 2

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You want that there exists an $\epsilon>0$, such that for all $\delta>0$, there exists $x$ and $y$ (depending on $\delta$) such that $|x-y| < \delta$, yet

$$ | f(x) - f(y) | > \epsilon. $$

Let's take $\epsilon = 1$.

Hint: suppose $x < y$. The MVT says there exists $c \in (x,y)$ such that

$$ | f(x ) - f(y) | = f'( c) | x - y |.$$ As $f'(c) = 3/2 c^{1/2} > 3/2 x^{1/2}$, one has that $$ | f(x ) - f(y) | > 3/2 \, x^{1/2}\, |x -y|.$$ (N.B. $x

In this approach, one must choose $x$, $y$, such that $|x-y| <\delta$, yet the right hand side is greater that $1$.

OK? I hope this helps!

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For fixed $k > 0$, let

\begin{align*} g(x) &= \sqrt{x^3}\\[4pt] h(x) &= g(x+k) - g(x) \end{align*}

If you can show that $h$ is increasing on $[1,\infty)$, that would imply $h(x) \ge h(1)$ for $x \ge 1$.

But you claimed $h(x) \ge h(1)$ for $x \ge1$ based solely on the fact that $g$ is increasing on $[1,\infty)$. That's not a valid justification.

Since g is increasing on $[1,\infty)$, it follows that

\begin{align*} g(x + k) &\ge g(1+k)\\[4pt] g(x) &\ge g(1) \end{align*}

for $x \ge 1$, but you can't subtract those inequalities.

Instead, to show h is increasing on $[1,\infty)$, show that $h^\prime(x) > 0\,$ on that domain.

Other than that, your proof looks OK.