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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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    @user26139 You would have: $(x_{n_k})$ converges to $x$. But for each $k$, $f (x_{n_k})\le{1\over n_k}$. Since ${1\over n_k}$ converges to 0 (since $n_k$ is strictly increasing) $f(x)=\lim_k f(x_{n_k})\le 0$.2012-04-15

2 Answers 2

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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 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 $x_{r_n}\to l\implies f(x_{r_n})\to f(l)\neq 0!$