Study the uniform continuity of $f(x) = \sqrt{x^2+x+1}$

Question

I started learning continuity and uniform continuity but the textbook I am following does not provide any example after explaining the uniform continuity. I have a function $f:\mathbb{R}\rightarrow \mathbb{R}; f(x) = \sqrt{x^2+x+1}$, and I need to study its uniform continuity. Would be nice if you could walk me through the whole process of solving this, not just solving it for me. Thank you!

Answer 1

Here is a very slow, very inelegant, very elementary solution, working directly and naively from the definition. It is meant to illustrate what kinds of choices are involved and what can go wrong in general.

Fix $x_0\in\mathbb{R}$. To prove continuity at $x_0$, we want to choose a neighborhood around $x_0$ on which the function values stay within $\epsilon$ of $f(x_0)$. Symbolically, we want to choose $\delta$ small enough so that

$$|x-x_0|<\delta \implies |f(x)-f(x_0)|<\epsilon$$

The trouble is that the inequality $|f(x)-f(x_0)|<\epsilon$ means

$$\left|\sqrt{x^2+x+1}-\sqrt{x_0^2+x_0+1}\right|<\epsilon$$

and this isn't the easiest thing to analyze. In particular, it's not immediately obvious how bounding $|x-x_0|$ will lead to a bound on the left-hand side.

One solution is to borrow an idea from the easiest proof that the square root function is continuous: just note that when $a$ and $b$ are nonnegative,

$$|\sqrt{a}-\sqrt{b}|\leq\sqrt{|a-b|}$$

But $x^2+x+1=(x+1/2)^2+3/4\geq0$, so we can apply this fact to our function. We get

$$\left|\sqrt{x^2+x+1}-\sqrt{x_0^2+x_0+1}\right|\leq\sqrt{\left|x^2+x+1-x_0^2-x_0-1\right|}=\sqrt{|x-x_0|}\sqrt{|x+x_0|+1}\tag{$\star$}$$

Now we are getting somewhere, because we can control $|x-x_0|$. But the term $|x+x_0|$ requires some trickery. (It's the same trickery required in the standard proof that $x\mapsto x^2$ is continuous.)

If we pick $|x-x_0|<\delta$, then $|x+x_0|=|x-x_0+2x_0|\leq|x-x_0|+2|x_0|\leq 2|x_0|+\delta$. It follows that $\sqrt{|x+x_0|+1}$ is bounded above by $\sqrt{2|x_0|+\delta+1}$.

What does this get us? If we ensure $\delta<1$, say, we will have bounded $\sqrt{|x+x_0|+1}$ by $\sqrt{2|x_0|+2}$. So, by ($\star$), $|f(x)-f(x_0)|$ will be bounded above by $\sqrt{\delta}\sqrt{2|x_0|+2}$. To ensure this is less than $\epsilon$, we pick $\delta=\epsilon^2(2|x_0|+2)^{-1}$. But our reasoning depended on $\delta$ being smaller than $1$, so to be safe we set $\delta=\min\left\{1,\epsilon^2(2|x_0|+2)^{-1}\right\}$.

At last, we have proven continuity for all $x_0\in\mathbb{R}$. Let us call this $\delta(x_0)$ to emphasize that the choice of $\delta$ depends on $x_0$.

Now, for uniform continuity, we need a $\delta$ that works for all $x_0$. The obvious candidate is the infimum of the $\delta(x_0)$ over $x_0\in\mathbb{R}$, but unfortunately this is zero: as $|x_0|\to\infty$, $\delta(x_0)\to0$. (However, if the domain of our function were a bounded interval rather than $\mathbb{R}$, we would be done, because the infimum of the deltas would be positive.) To fix this problem, we must see whether we can find another $\delta$ to use when $|x_0|$ is big.

For this, you can use M. Hallgren's approach: rationalize the numerator. We have

$$\left|\sqrt{x^2+x+1}-\sqrt{x_0^2+x_0+1}\right|=\left|\frac{x^2-x_0^2+x-x_0}{\sqrt{x^2+x+1}+\sqrt{x_0^2+x_0+1}}\right|\leq\frac{|x^2-x_0^2|+|x-x_0|}{\sqrt{x^2+x+1}+\sqrt{x_0^2+x_0+1}}=\frac{|x-x_0|(1+|x+x_0|)}{\sqrt{x^2+x+1}+\sqrt{x_0^2+x_0+1}}\leq\frac{|x-x_0|(1+|x+x_0|)}{|x|+|x_0|}\leq\frac{|x-x_0|\cdot 2|x+x_0|}{|x|+|x_0|}\leq2|x-x_0|$$

where the penultimate step follows for $|x_0|>2$ and $\delta<1$, because in that case $1\leq|x+x_0|$. So when $|x_0|$ is big (bigger than 2), we can now just pick $\delta=\min\left\{1,\epsilon/2\right\}$ -- which doesn't depend on $x_0$.

This shows that your function is uniformly continuous on all of $\mathbb{R}$.

The upshot of this tedious explanation is that it shows you what keeps a continuous function from being uniformly continuous: the infimum of the local deltas is zero, and there is no fix available, as there was in this case. Continuity tells us that we can control the variation of the function locally: around every fixed $x_0$ we can find an interval of radius $\delta(x_0)$ on which $|f(x)-f(y)|<2\epsilon$, no matter how small the $\epsilon$. (I say $2\epsilon$ because if $f(x)$ and $f(y)$ are both within $\epsilon$ of $f(x_0)$, they must be within $2\epsilon$ of each other.) For uniform continuity, we need a $\delta$ that works globally, no matter where the interval of radius $\delta$ is centered. It is a nice exercise to prove carefully that we can find such a $\delta$ if and only if the infimum of the function giving the best $\delta$ at each $x_0$ is positive.

Answer 2

It is clear that $|f'(x)| \le 1$ on $\Bbb R$ (it is increasing and goes to $\pm 1$ when $x \to \pm \infty$). Hence, for all $x$ and $y$,

$$|f(x) - f(y)| = \left| \int_y^x f'(t) dt \right| = \left| \int_{\min(x,y)}^{\max(x,y)} f'(t)dt \right| \le \int_{\min(x,y)}^{\max(x,y)} |f'(t)|dt \le |x - y|$$

Choosing $\delta = \epsilon$ does the job.

Answer 3

We can prove the inequality $$|f(x) - f(y)| = |\sqrt{x^2 + x + 1} - \sqrt{y^2 + y + 1}| \leq |x-y|$$ directly. Observe that $$\begin{aligned} & \qquad \quad |\sqrt{x^2 + x + 1} - \sqrt{y^2 + y + 1}| \leq |x-y| \\ & \iff (\sqrt{x^2 + x + 1} - \sqrt{y^2 + y + 1})^2 \leq (x-y)^2 \\ &\iff (x^2 + x + 1) - 2\sqrt{(x^2 + x + 1)(y^2 + y + 1)} + (y^2 + y + 1) \leq (x-y)^2 \\ &\iff x + y + 2xy + 2 \leq 2\sqrt{(x^2 + x + 1)(y^2 + y + 1)} \\ \end{aligned}$$ Now observe that the inequality $$(x + y + 2xy + 2)^2 \leq 4(x^2 + x + 1)(y^2 + y + 1)$$ is true, as can be seen by expanding both sides and reducing to $$0 \leq 3(x-y)^2$$ Consequently, taking square roots, we get $$|x + y + 2xy + 2| \leq 2\sqrt{(x^2 + x + 1)(y^2 + y + 1)}$$ so certainly $$x + y + 2xy + 2 \leq 2\sqrt{(x^2 + x + 1)(y^2 + y + 1)}$$ which is the last inequality in the chain of $\iff$'s above. Consequently the first inequality in the chain is also true.

Answer 4

You can write $$\left| \sqrt{x^2 + x + 1} - \sqrt{y^2 + y + 1} \right| = \left| \dfrac{(x^2 - y^2) + (x - y) }{\sqrt{x^2 + x + 1} + \sqrt{y^2 + y + 1}} \right| \leq \left| \dfrac{(x^2 - y^2) + (x - y) }{\sqrt{x^2 + x + 1} + \sqrt{y^2 + y + 1}} \right| \\ \leq \dfrac{(1 + |x + y|) \cdot |x - y|}{\left| \sqrt{x^2 + x + 1} + \sqrt{y^2 + y + 1} \right|} . $$ To get a upper bound on this term, we want a lower bound on the denominator. Least squares gives $x^2 + x + 1 = (x + 1/2)^2 + 3/4 \geq 3/4$, and similarly for $y$, so we obtain $$|\sqrt{x^2 + x + 1} + \sqrt{y^2 + y + 1}| \geq 2\sqrt{3/4} = \sqrt{3}. $$ However, this lower bound is not good enough for when $|x|$ or $|y|$ is large, so we restrict to the case when either $|x| \geq 2$ or $|y| \geq 2$ (assume without loss of generality that $|x| \geq 2$). Then $$x^2 + x + 1 \geq \frac{1}{2}x^2 + (\frac{1}{2}x^2 -|x|) = \frac{1}{2}x^2 + \frac{1}{2}|x|(|x| - 2) \geq \frac{1}{2}x^2,$$ and so $\sqrt{x^2 + x + 1} \geq \frac{1}{\sqrt{2}}|x|$. The same estimate holds when $y$ is large, and you should be able to fill in the details from there. This shows that your function is in fact Lipschitz.