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If $g(x):= \sqrt x $ for $x \in [0,1]$, show that there does not exist a constant $K$ such that $|g(x)| \leq K|x|$ $ \forall x \in [0,1]$

Conclude that the uniformly continuous function $g$ is not a Lipschitz function on interval $[0,1]$.

Necessary definitions:

Let $A \subseteq \Bbb R$. A function $f: A \to \Bbb R$ is uniformly continuous when: Given $\epsilon > 0$ and $u \in A$ there is a $\delta(\epsilon, u) > 0$ such that $ \forall x \in A$ and $|x - u| < \delta(\epsilon,u)$ $\implies$ $|f(x) - f(u)| < \epsilon$

A function $f$ is considered Lipschitz if $ \exists$ a constant $K > 0$ such that $ \forall x,u \in A$ $|f(x) - f(u)| \leq K|x-u|$.

Here is the beginning of my proof, I am having some difficulty showing that such a constant does not exist. Intuitively it makes sense however showing this geometrically evades me.

Proof (attempt):

Suppose $g(x): = \sqrt x$ for $x \ in [0,1]$ Assume $g(x)$ is Lipschitz. $g(x)$ Lipschitz $\implies$ $\exists$ constant $K > 0$ such that $|f(x) - f(u)| \leq K|x-u|$ $\forall x,u \in [0,1]$.

Evaluating geometrically:

$\frac{|f(x) - f(u)|}{ |x-u|}$ = $\frac{ \sqrt x - 1}{|x-u|}$ $ \leq K$

I was hoping to assume the function is Lipschitz and encounter a contradiction however this is where I'm stuck.

Can anyone nudge me in the right direction?

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  • $\begingroup$ HINT: The inequality you want to disprove can be rephrased as follows: "the function $|x|^{-\frac12}$ is bounded on $(0,1]$". Is this true? If you can show it's not, you are finished. $\endgroup$ – Giuseppe Negro Feb 2 '16 at 23:37
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Take $u=0$. Then you should have $|f(x)|\le K|x|$ for $x\ne0$. Notice that $|f(x)|/|x|\to\infty$ when $x\to0$, which contradicts to this inequality.

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Well, choose $u=0$. What can you say about $\frac{|f(x)-f(u)|}{|x-u|}=\frac{\sqrt{x}}{x}$ as $x$ gets closer to zero?

Here is another proof using derivatives.

Since $f'(x)=\frac{1}{2\sqrt{x}}$, it follows that $$\lim_{x\to 0} f'(x)=+\infty$$ This means that for every $M>0$ there exists $\delta$ such that for every $x$ in $I=[0, \delta]$, $f'(x)>M$. Suppose by contradiction that such $K$ exists. Consider $M=K+1$ and $x_0\in \text{int}(I)$: then $$f'(x_0)=\lim_{x\to x_0}\frac{f(x)-f(x_0)}{x-x_0}>K+1$$ But then there exists an interval $J$ such that for every $x\in I \cap J$

$$\frac{f(x)-f(x_0)}{x-x_0}>K+1$$ $$\implies |f(x)-f(x_0)|>(K+1)|x-x_0|$$

contradicting our hypothesis.

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