# Continuous Function Greater than 0

Let $f$ be continuous on $[a, b]$. Suppose $f(x) > 0$ for all $x \in [a, b]$. I'm trying to show that there exists a $\alpha > 0$ such that $f(x) > \alpha$ for all $x \in [a, b]$.

I tried to prove this by contradiction.

Assume that for every $\alpha > 0$, there exists an $x \in [a, b]$ such that $f(x) \leq \alpha$. Then I let $\alpha_n = \frac{1}{n} > 0$. Then there exists an $x_n \in [a, b]$ such that $f(x_n) \leq \alpha_n$. But note that $\alpha_n \to 0$ as $n \to \infty$. This implies that there is an $x_n$ such that $f(x_n) \leq 0$, which is a contradiction.

Could someone give me feedback on my proof?

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Your intuition is correct. However you can't say that $f(x_n)\leq 0$ for some $n$. If $f(x_n)=\frac {1}{n}$ then it is obviously false. – azarel Apr 15 '12 at 19:44
Three observations: (1) You have not used continuity of $f$. (2) "This implies that..." is not correct. (3) Do you know, and can you use, that a continuous function defined on $[a,b]$ attains its minimum? – Xabier Domínguez Apr 15 '12 at 19:45
The proof is not right. Something like it can be made to work. You need some properties of real numbers. I don't know which ones have been proved for you at this stage. Has it been proved already that a bounded sequence has a convergent subsequence? – André Nicolas Apr 15 '12 at 19:46
@AndréNicolas: I have learned about the Bolzano-Weierstrass Theorem. By the theorem $\{x_n\}$ has a convergent subsequence namely $\{x_{n_k}\}$. Then $f(x_{n_k}) \leq \alpha$. But should $\alpha = \frac{1}{k}$ or $\alpha = \frac{1}{n_k}$??? – user26139 Apr 15 '12 at 19:57
@user26139 You'd use $f(x_{n_k})\le {1\over n_k}$. (the other choice would work too). But, note that it's the limit of this subsequence that leads to the contradiction. – David Mitra Apr 15 '12 at 20:04

It's not correct. Where do you get this $x_n$?

But, by the compactness of $[a,b]$, your argument can be salvaged. There is a subsequence of $(x_n)$ that converges to an $x\in[a,b]$; and by the continuity of $f$, we would have $f(x)\le0$.

Or, arguing directly, you could consider the minimum value of $f$ on $[a,b]$ (which exists, since $[a,b]$ is closed and bounded).

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For otherwise $f$ becomes arbitrarily close to $0.$ Consequently $\exists$ a sequence $\{x_n\}$ in $[a,b]$ such that $$0<f(x_n)<\frac{1}{n}$$ Squeezing $f(x_n)\to 0.$

Since $\{x_n\}$ is bounded $\exists$ a convergent subsequence $\{x_{r_n}\}$ of $\{x_n\}$ converging to some $l\in[a,b].$

$f$ is continuous at $l$ and $x_{r_n}\to l\implies f(x_{r_n})\to f(l)\neq 0!$

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