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Let $p(x)$ be a non-constant polynomial with real coefficients such that $p(x) \neq 0$, $\forall x \in \mathbb{R}$. Define $f(x)= \frac{1}{p(x)}$ for all real $x$. Prove that

  1. For each $\epsilon >0$ there exists a $\delta >0$ such that $|f(x)| < \epsilon $ for all real $x$ satisfying $|x| > \alpha$,
  2. $f: \mathbb{R} \to \mathbb{R}$ is a uniformly continuous function.

Attempt:

  1. $p(x)$ being unbounded above and being ultimately monotone (highest power dominates), ( WLOG, we have assumed the polynomial is $>0$), $\forall G>0$, $\ \exists x_0 \in \mathbb{R}$, s.t. $p(x)>G$ whenever $x> x_0$ and $x< -x_0$ [Highest power even]. Choosing $G$ with $1/G< \epsilon$, we get $1/p(x) < 1/G < \epsilon$ whenever $|x|> x_0$.

2: We show that $f'(x)$ is bounded for every $x$.

$f(x)= \frac{1}{p(x)} \implies f'(x) = \frac{-p'(x)}{(p(x))^2}$. Now, $p(x)$ is either $>0$ for $<0$ for all $x$. [No change of sign can occur].

Clearly $|p(x)|= \mathcal{O}(x^{2m})$. $|p'(x)| =\mathcal{O}(x^{2m-1})$. Hence, $|f'(x)|= \mathcal{O}(\frac{1}{x^{2m+1}}) < M$ for some $M \in \mathbb{R}^{+}$.

Is this correct? Please justify.

Thank You.

Subhasis Biswas
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  • It is fine. You can try now proving that if a function $\mathbb{R}\to\mathbb{R}$ is continuous and it tends to $0$ as $x\to\pm\infty$, then it is uniformly continuous. Solving this one will give you a new proof that doesn't require the existence or use the derivative. – user647486 Apr 24 '19 at 14:58
  • @user647486 okay! I will try to prove this one. – Subhasis Biswas Apr 24 '19 at 15:02
  • Is there a missing $\exists \alpha$? – Acccumulation Apr 24 '19 at 15:04
  • By the way, the implication that an element of $\mathcal{O}(x^{2m-1})$ and an element of $\mathcal{O}(x^{2m})$ is an $\mathcal{O}\left(\frac{1}{x^{2m+1}}\right)$ doesn't follow in general. Not even $\mathcal{O}(1/x)$ or $\mathcal{O}(1)$. – user647486 Apr 24 '19 at 15:06
  • But what about this specific case? – Subhasis Biswas Apr 24 '19 at 15:53

1 Answers1

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I guess that the expected answer should use the first question of the exercise. And indeed you can conclude easily knowing the first result : take $\varepsilon>0$, then outside your $]-x_0, x_0[$, $f$ does respect the inequality of the uniform continuity for a good $\delta$. But $f$ is uniformly continuous on $[-x_0-1, x_0+1]$ by Heine theorem, because this is a compact segment. You can conclude given all that.

Note that your solution does work, as explained in this thread.

elidiot
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  • Yes, I guess it is indeed a follow-up question. I was trying to figure out a suitable $\delta$. But then I went around it. – Subhasis Biswas Apr 24 '19 at 15:00
  • Is my method good enough (both 2nd and the first problem)? – Subhasis Biswas Apr 24 '19 at 15:01
  • It is fine for the first question, and the second question is also ok if you conclude with the mean value theorem. But do try to conclude by hand as the exercise was expected, this is a good exercise in itself. – elidiot Apr 24 '19 at 15:04
  • The link is very helpful. Surprisingly, an answer uses the exact same approach. The thing is, the function is very much well behaved (as pointed out in the comments) . – Subhasis Biswas Apr 24 '19 at 15:05
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    Indeed, and your solution couldn't be adapted to the case where $P$ is replaced by any continuous function $f>0$ such that $f(x)\to+\infty$ when $x\to\pm\infty$, which is still true. – elidiot Apr 24 '19 at 15:07