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Show $\frac{1}{n^{0.5}}$ is continuous for $[1,\infty]$.

I am unsure how to go about showing this, anyone have any ideas?

user2250537
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3 Answers3

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Let us show that $f:x\mapsto x^{-1/2}$ is continous on $[1,+\infty]$ . First, $f_{1}:x\mapsto x^{1/2}$ is continous on the same interval : indeed, if $x,y\geq1$ are real numbers with and $|x-y|<\delta$ , then we have$$\left|f_{1}\left(x\right)-f_{1}\left(y\right)\right|=\left|x^{1/2}-y^{1/2}\right|=\frac{\left|x-y\right|}{x^{1/2}+y^{1/2}}<\frac{\delta}{2}$$ since $x\geq1$ and $y\geq1$ .

Now, let us consider the function $f_{2}:x\mapsto x^{-1}$ on the same interval : for $x,y$ as above, we have$$\left|f_{2}\left(x\right)-f_{2}\left(y\right)\right|=\left|\frac{1}{x}-\frac{1}{y}\right|=\left|\frac{y-x}{xy}\right|=\frac{\left|y-x\right|}{xy}\leq\left|y-x\right|<\delta$$ since $1\leq xy$ .

Now, since $f\left(x\right)=f_{2}\left(f_{1}\left(x\right)\right)$, we use the fact that a composition of continuous functions is also continuous, and then we get the continuity of $f$ .

hausdork
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Nicolas
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    the way you show is uniformly continuos, and is it supposed to be less that $\epsilon$ not $\delta$ – Alexander Mar 24 '15 at 18:54
  • What I wrote shows the continuity at $x$ for any $x$. Now, if I fix $\varepsilon>0$, I can find $\delta>0$ such that $$\left(\left|x-y\right|<\delta\right)\Longrightarrow\left(\left|f\left(x\right)-f\left(y\right)\right|<\varepsilon\right)$$ just by taking $\delta=\varepsilon$. – Nicolas Mar 24 '15 at 19:09
  • oh, in my text book, the way you show is uniformly continuous, not continuous. Anyway, uniformly is strong that continuous by itself. – Alexander Mar 24 '15 at 19:15
  • Take care of something : I define $\delta$ as above for a certain $x$, so maybe $\delta$ depends on $x$. The uniform continuity is to find a uniform constant $\delta$ for any $x$ of the interval. – Nicolas Mar 24 '15 at 19:23
  • @Nicolas Good point, your proof stands. – Paddling Ghost Mar 24 '15 at 19:48
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Let $x,\; x_0 \in [1, \infty]$, we have: $$\left| \frac{1}{\sqrt{x}} - \frac{1}{\sqrt{x_0}} \right| = \left| \frac{\sqrt{x_0} - \sqrt{x}}{\sqrt{x}\cdot\sqrt{x_0}}\right| $$ $$< \left |\frac{x - x_0}{1} \right|$$ $$= |x - x_0| < \delta$$ Let $\delta = \epsilon$

Then by definition of continuity, for each $\epsilon > 0$ there exists $\delta > 0$ so that $x \in [1,\infty]$ and $|x-x_0| < \delta$ imply $|f(x) - f(x_0)| < \delta = \epsilon$

Alexander
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  • This works, but showing $\sqrt x_0 - \sqrt x \le x-x_0$ is not trivial. Indeed, it only works for our specific case because $x,x_0 \ge 1$. Still, good proof and the OP should be able to find that step in one of the other answers. – Paddling Ghost Mar 24 '15 at 19:46
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Following the hint provided by copper.hat, we will show that the function $f(x)=\sqrt x$ is continuous on the interval $[1,\infty)$ and then show that the function $g(x) = 1/x$ is continuous on the same interval. Recall the definition of a continuous function on a set $A \subset \mathbb{R}$: $\forall \epsilon >0$, $\exists \delta >0$ and $\forall a \in A$ if $|x-a| < \delta$, then $|f(x)-f(a)| < \epsilon$.

1) $\sqrt x $ is continuous on A. We look at the difference $|\sqrt x - \sqrt a|$. Since $(x-y) = (\sqrt x + \sqrt a)(\sqrt x - \sqrt a)$ we have that $|\sqrt x - \sqrt a| = \frac{|x-a|}{|\sqrt x + \sqrt a|} \le \frac{\delta}{|\sqrt x + \sqrt a|}$. Since $x \ge 1 $ and $a \ge 1$, then $ |\sqrt x + \sqrt a| \ge 2$. Thus, $ \frac{\delta}{|\sqrt x + \sqrt a|} \le \delta / 2$. Let $\delta = 2\epsilon -1$, then $\frac{\delta}{|\sqrt x + \sqrt a|} \le \delta / 2 = \epsilon - 1/2 < \epsilon$.

2) $\frac{1}{x}$ is continuous on A. This will be proved similarly. Again, consider the difference $|\frac{1}{x}-\frac{1}{a}|$. $$|\frac{1}{x}-\frac{1}{a}|= \frac{|a-x|}{|ax|}=\frac{|x-a|}{ax}$$ (we can ommit the absolute value around the bottom since the product of two positive numbers is surely positive.) Now we have that $|x-a|< \delta$, and, again, since $x \ge 1,y\ge 1$, $ax \ge 1$. Thus, $$\frac{|x-a|}{ax} \le \delta$$ Now choose $\delta = \epsilon -1$ to arrive at $\frac{|x-a|}{ax} < \epsilon$.

3) Fact: the composition of two continuous functions is a continuous function. (If we wanted to be really really rigorous we would have to prove this ourselves.) Since, $$\frac{1}{\sqrt x} = g(f(x))$$ ,where $g(x)=\frac{1}{x}$ and $f(x) = \sqrt x$, $\frac{1}{\sqrt x}$ is continuous on A.

Paddling Ghost
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