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From Spivak's book

Suppose $f_n = e^{-x^2}/n$ be defined on $\mathbb{R}$. Is $(f_n)$ uniformly convergent?

Spivak uses the M-Weistress Test to show that it is uniformly convergent. I did something different and showed that it is not uniformly convergent.

So I did $|f_n - f| = |e^{-x^2}/n - 0| = |e^{-x^2}/n| = e^{-x^2}/n < \epsilon \iff n > \epsilon/e^{-x^2}$

So there is no one integer $N$ that works for every $x$. What did I do wrong?

EDIT: From Spivak's answer, taken from chapter 23 problem 1v. (Unfortunately I could not locate the ed of my book, but I believe it is the 2nd as the key I have is the 3rd ed and is one chapter ahead of the book I use)

(v) $f(x) = \lim_{n \to \infty} f_n(x) = 0$, and $\{ f_n\}$ converges uniformly to $f$, since $| f_n(x) | \leq 1/n$ for all $x$

Lemon
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    I am skeptical about the statement that Spivak uses the Weierstrass M-test for this. Could you please give a precise reference? – Jonas Meyer Jan 06 '13 at 06:09
  • He didn't state the test, but he used it. Unfortunately this is one of the few selected answers in the book and not in the answer book, so I'll type it out and hope that you believe me – Lemon Jan 06 '13 at 06:17
  • sizz: I don't understand. If it is in a book, could you say what book and on what page? – Jonas Meyer Jan 06 '13 at 06:19
  • See edit please thanks – Lemon Jan 06 '13 at 06:22
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    sizz: Thanks very much. The Weierstrass M-test is a criterion for convergence of series. Here what is being used is that $f_n\to g$ uniformly if there is a sequence $(a_n)$ of positive numbers such that $a_n\to 0$ and $|f_n(x)-g(x)|\leq a_n$ for all $x$ and all $n$. (The "only if" direction would also hold, but isn't needed here.) I don't think that this criterion has a common name. – Jonas Meyer Jan 06 '13 at 06:25
  • Ah you are right, I just checked because $\sum M_n = \sum 1/n$ is the harmonic series. Okay so what he did just resembled what I thought it was. Thanks – Lemon Jan 06 '13 at 06:29
  • A related problem. – Mhenni Benghorbal Jan 06 '13 at 06:54

2 Answers2

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Note that $\forall x\in\mathbb{R}[e^{-x^2}\leq 1]$, thus : $$\forall n>\frac{1}{\epsilon}[|\frac{e^{-x^2}}{n}|<\epsilon]$$

Amr
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Amr shows why the convergence is uniform. I think that an answer to your question is that you made an algebra error. For positive numbers $a,b,c$, the inequality $a/b <c$ is equivalent to $b>a/c$, not $b>c/a$.

Jonas Meyer
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  • Regardless my answer still would've been dependent on $x$. But thank you for pointing it out – Lemon Jan 06 '13 at 06:08
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    @sizz: Your answer would have been not done yet, but it would have had hope. As you pointed out, there is no way to choose $n$ (depending on $\varepsilon$) so that $n>\varepsilon/e^{-x^2}$ for all $x$, because $e^{-x^2}$ can be arbirarily small. However, it is easy to find $n$ (depending on $\varepsilon$) such that $n>e^{-x^2}/\varepsilon$ for all $x$. – Jonas Meyer Jan 06 '13 at 06:11