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can I have the rationale for the first line of this proof? i.e. How did you know to start answering the question in this manner? I am guessing it is because you want to exploit the definition of continuity? (I am able to follow the rest of the proof, no problem)

The question goes like this:

Suppose that $f: \mathbb{R} ^{m} \rightarrow \mathbb{R} $ is a continuous function and that $f(x^{*}) \gt 0$. Show that there is an open ball B = $B_{\delta}(x^{*})$ such that $f(x) \gt 0$ for all x $\epsilon$ B.

Solution:

Let $\epsilon$ be such that $0 \lt \epsilon \lt f(x^{*})$

By continuity of f at $x^{*}$, there is a $\delta \gt 0$ such that $|| x - x^{*} || \lt \delta \Rightarrow |f(x) - f(x^{*})| \lt \epsilon $

But $|f(x) - f(x^{*})| \lt \epsilon $ implies $f(x) \gt f(x^{*}) - \epsilon \gt 0$

So for all x satisfying $|| x - x^{*} || \lt \delta$, $f(x) > 0$

But this is simply saying that for all x $\epsilon B = B_{\delta}(x^{*}), f(x^{*}) \gt 0$

My question:

How did you know to start the proof using this line:

Let $\epsilon$ be such that $0 \lt \epsilon \lt f(x^{*})$?

Many thanks

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You're pretty much right on with seeing the connection between the first line and the continuity condition. Looking at this problem, we see that the property we're looking for is something local - that is, we're only interested in what happens near the point $x^*$. Continuity is also a local property, so we're on the right track. Furthermore, all we know about the function $f$ is that it's continuous, so we want to use that somehow. In fact, we can look directly at we want to prove: that, given something away from 0, we can keep it away from 0 on some small neighborhood. All of this is to say, everything about this problem points to continuity, so we're going to directly use the definition of continuity from the beginning.

Figuring out that we want $\epsilon < f(x^*)$ comes from a bit of work with the function - we want something small enough such that $|f(x^*) - f(x)| < \epsilon \Rightarrow f(x) > 0$. The first inequality tells us $f(x) > f(x^*) - \epsilon$, so the choice of $\epsilon$ follows almost immediately from this. Hope this helps!

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  • $\begingroup$ Thank you Zachs, this is very clear. Can you please elaborate a little more on this line "ϵ<f(x∗) comes from a bit of work with the function" - as in what else is needed? Because if we have already assumed f(x∗) > 0, then as how we can make ϵ as small as we want to be, is this simply not taking ϵ < f(x*) ? $\endgroup$ – beginner Apr 22 '15 at 7:26
  • $\begingroup$ Yes, that's exactly what that means. I just meant that it wouldn't be immediately obvious that that would be the choice of $\epsilon$ without thinking about it for a bit. $\endgroup$ – Zach Kirsche Apr 23 '15 at 2:37
  • $\begingroup$ Thanks again Zach, just wanted to be sure on that point. $\endgroup$ – beginner Apr 23 '15 at 3:17
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Assume $f$ as a function from $\mathbb{R}$ to $\mathbb{R}$ and try to visualize the proof.

To prove $f(x) \gt 0 \forall x \in B_\delta(x^*)$ for some $\delta$,

we want to prove the that $f(x) \in B_\epsilon(f(x^*))$ for some $\epsilon$ and $\forall x \in B_\delta(x^*)$.

But arbitrary $B_\epsilon(f(x^*))$ can contain zero.

So in the proof we choose $\epsilon$ such a way that $0 \notin B_\epsilon(f(x^*))$ .

once you have such an $\epsilon$, proof immediately follows from the definition of continuity.

note that $f$ is continuous at $x^*$ iff given $\epsilon \gt 0$ there exists $\delta \gt 0$ such that $f(B_\delta(x^*)) \subset B_\epsilon(f(x^*)).$

I hope this justifies why we start with such an $\epsilon$.

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  • $\begingroup$ Thank you GA316, I wish I could accept more answers $\endgroup$ – beginner Apr 22 '15 at 7:30

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