Proof that $f(x)=\frac{1}{1+x^4}$ is continuous

Question

Im struggling with the proof that $$f(x)=\frac{1}{1+x^4}$$ is continuous. I want to use the epsilon-delta definition. For $$\frac{1}{1+x^2}$$ there is a proof in this forum, but for $$\frac{1}{1+x^4}$$ its more complicated I guess.

I didn't find it yet (didn't look very hard tbh) but intuitively you should be able to make that work by replacing $x \mapsto x^2$, i.e. given $\epsilon > 0$ choose $\delta_0 > 0$ such that $|x^2 - a| < \delta_0 \implies |f(x) - f(a)| < \epsilon$ , then reason back what your actual $\delta$ should be in terms of $\delta_0$. — CompuChip, Jan 04 '17 at 08:18
http://math.stackexchange.com/questions/997602/show-1-1-x2-is-uniformly-continuous-on-bbb-r — KCK, Jan 04 '17 at 08:19
@KCK: You said that 'its more complicated i guess'.Why don't you write down and show us where are you facing problem? — Arpit Kansal, Jan 04 '17 at 08:28
@KCK Factoring $x-y$ out of $x^4-y^4$ is doable using long division. — Hetebrij, Jan 04 '17 at 09:00
yes, I did that, but than i did not not know how to handle the remaining polynomial (see your answer!). — KCK, Jan 04 '17 at 09:06
$1/y$ has only one not-continuous point, y=0. However, there are no real number such $1+x^4=0$. Thus, $1/(1+x^4)$ is continuous. — pasaba por aqui, Jan 04 '17 at 15:32

score 4 · Answer 1 · edited Apr 13 '17 at 12:20

We have for $x,y \in \mathbb{R}$ \begin{align} \left| \frac{1}{1+x^4} - \frac{1}{1+y^4} \right| &= \left| \frac{y^4 - x^4}{(1+x^4)(1+y^4)} \right| \\ &= \left| \frac{(x-y)(x^3+x^2y+xy^2+y^3)}{(1+x^4)(1+y^4)} \right| \\ &=|x-y| \left| \frac{x^3+x^2y+xy^2+y^3}{(1+x^4)(1+y^4)} \right| \\ &= |x-y| |x+y|\left| \frac{x^2+y^2}{(1+x^4)(1+y^4)} \right| \end{align} So from the proof that $\frac{1}{1+x^2}$ is continuous, we have that $\left| \frac{x^2+y^2}{(1+x^4)(1+y^4)} \right| \le 1$, replace $x_1$ and $x_2$ by $x^2$ and $y^2$.

So given $x \in \mathbb{R}$ and $\epsilon >0$, if $|x-y| < \delta$, then \begin{align} \left| \frac{1}{1+x^4} - \frac{1}{1+y^4} \right| \le |x-y| |x+y| \le \|x-y|(2|x| + |x-y|) \le \delta (2|x| + \delta) = \epsilon \end{align} provided we choose $\delta = \sqrt{|x|^2+\epsilon}-|x|$.

thank you! however, since the question appeared in a basic text by Courant: Isn't there a shorter/"easier" solution? — KCK, Jan 04 '17 at 09:09

score 2 · Accepted Answer · answered Jan 04 '17 at 09:15

2

Assume WLG that $|y|\geq|x|$. Then $$\left|{1\over 1+x^4}-{1\over 1+y^4}\right|=|x-y|\>{\bigl|x^3+x^2y+xy^2+y^3\bigr|\over(1+x^4)(1+y^4)}\leq|x-y|\>{4|y|^3\over1+y^4}\ .$$ It is easy to see that ${4|y|^3\over1+y^4}\leq C$ for some $C>0$. It follows that the function $f$ in question is even Lipschitz continuous with Lipschitz constant $C$. In other words: You can take $\delta:={\epsilon\over C}$ througout.

answered Jan 04 '17 at 09:15

Christian Blatter

226,825

why can you assume WLG that y>x? what about x>y, then the above inequality does not work? – KCK Jan 04 '17 at 17:28
Assuming $|y|\geq|x|$ is notationally simpler than introducing a new variable $r:=\max{|x|,|y|}$. – Christian Blatter Jan 04 '17 at 18:42

score 1 · Answer 3 · answered Jan 04 '17 at 07:58

1

$f$ can be written in form $P/Q$ where $P=1$ is trivially continuous and $Q=x^4+1$ is continuous since it is polynomial.

Then it is enough to show that $Q$ is always non-zero.

answered Jan 04 '17 at 07:58

Will Sherwood

1,371

1

Yes. But what about using just the delta-epsilon definition? – KCK Jan 04 '17 at 08:03

score 0 · Answer 4 · answered Jan 04 '17 at 15:22

HINT.-If you have no problem with the continuity $\dfrac{1}{1+x^2}$ and you have problem with bounding $\left|\dfrac{y^4-x^4}{x+y)(x^2+y^2) }\right|$ then you can use the following identity:

$ $ $$\frac{1}{1+x^4}=\frac{1}{2\sqrt2}\left(\frac{x+\sqrt2}{1+\sqrt2x+x^2}-\frac{x-\sqrt2}{1-\sqrt2x+x^2}\right)$$

Proof that $f(x)=\frac{1}{1+x^4}$ is continuous

4 Answers4