Proof for Mendelson 2.4.8 - preserving positivity under continuity

Intro

I recently finished a proof for a problem that I found particularly interesting. The problem itself is about proving that if a continuous function $f: \mathbb{R} \rightarrow \mathbb{R}$ is positive at some point $a$, then there’s a closed interval $F$ around $a$ on which $f$ takes only positive values. In fact, there is a positive constant $k$ such that every value of $f$ on $F$ is at least $k$. Or in other words, $k$ is a positive lower bound for $f$ on $F$.

I like this problem for multiple reasons. First, the result is interesting. Informally, it means that continuity of $f:\mathbb{R} \rightarrow \mathbb{R}$ implies we can select a neighborhood of $a$ that preserves a property of $f(a)$ itself. In this case, that property is positivity. Given the density of real numbers, this result feels intuitive to me, but it’s still nice to see it proven formally.

Additionally, working through the problem was both challenging and fun. It forced me to invoke continuity in a way that was new to me, in that I had to come up with an $\epsilon$ and a $\delta$! You might say it forced me to use the claw of the hammer. Additionally, my first attempt at the proof—which I will share—was wrong; I chose an $\epsilon$ that was a little too slippery and ended up importing heavy machinery for a light problem.

Anyway, here’s the problem, formally stated.

The problem

Let $\mathbb{R}$ be the set of real numbers and $f:\mathbb{R} \rightarrow \mathbb{R}$ a continuous function. Suppose that for some $a \in \mathbb{R}$, $f(a) \gt 0$. Prove that there is a positive number $k$ and a closed interval $F = [a - \delta, a + \delta]$ for some $\delta \gt 0$ such that $f(x) \geq k$ for $x \in F$.

Proof - attempt 1

Let $\epsilon = f(a)$. Since $f$ is continuous, we may take $\delta'$ such that $|f(x) - f(a)| \lt \epsilon$ whenever $|x - a| \lt \delta'$. Choose $\delta = \frac{\delta'}{2}$. Let $F = [a - \delta, a + \delta]$ and $k = \min{f(F)}$. Since $|x - a| \leq \delta \lt \delta'$ for all $x \in F$, we have $|f(x) - f(a)| \lt f(a)$. Therefore $f(x) \gt 0$. Since $k = \min{f(F)}$, it follows that $k \gt 0$ and $f(x) \geq k$, as required.

Reflection

The problem with this proof is the choice of $k$. It quietly assumes that the image of the closed interval $F$ contains a minimal value. In other words, it makes use of the Extreme Value Theorem. That’s not material I have covered in Mendelson.

Beyond that, if we’re smarter with our choices of $\epsilon$ and $k$, we can avoid the need for such a heavy tool.

Proof - attempt 2

Let $\epsilon = \frac{f(a)}{2}$. Since $f$ is continuous, we may take $\delta'$ such that $|f(x) - f(a)| \lt \epsilon$ whenever $|x - a| \lt \delta'$. Choose $\delta = \frac{\delta'}{2}$. Let $F = [a - \delta, a + \delta]$ and $k = \epsilon$. Since $|x - a| \leq \delta \lt \delta'$ for all $x \in F$, we have $|f(x) - f(a)| \lt \frac{f(a)}{2}$. Hence $f(x) \gt \frac{f(a)}{2} = k$, as required.

Reflection

We end up proving a stronger statement than what the problem asks for, since we demonstrate $f(x) \gt k$. And I think that’s a nice takeaway. Prove the stronger statement if it’s easier.

But the real lesson here likely has to do with sharpening my sense for when something is justified. To me, taking $k = \min{f(F)}$ seemed intuitive. I implicitly reasoned that a continuous function on a closed interval should attain a minimal value on that interval.

The problem is that my intuition was leaning on ideas that I had not yet earned the right to use.