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I'm a bit confused about what my professor mentioned about the continuity of functions. Let $f$ be continuous on $[a, b]$, and suppose there exists a $c \in [a, b]$ such that $f(c) > 0$. Then there exists a $\delta$ such that $f(x) > 0$ for $|x-c| < \delta$.

Why is this the case? (This is not homework).

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Don't look away until you understand it :-) Now suppose you are looking at a moving target... But seriously, it helps to know what it is not: not a jump, and not an infinitely fast oscillation, etc. –  bgins Apr 18 '12 at 19:10
    
The idea is that if $f$ is continuous and positive for a value $x=c$, then it must be the case that for certain $\delta >0$, must musn't change sign in the neighborhood of $c$ which we tend to express as $|x-c|<\delta$ –  Pedro Tamaroff Apr 18 '12 at 19:10
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As Beni is suggesting, think about $|f(x)-f(c)|<\epsilon$ What happens if we choose $\epsilon = f(c)$? –  Pedro Tamaroff Apr 18 '12 at 19:12
    
@PeterT.off: We get $0 < f(x)$, which is what my professor said. –  user26139 Apr 18 '12 at 19:18
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Precisely, so that $f>0$ under that conditions. –  Pedro Tamaroff Apr 18 '12 at 19:28
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4 Answers

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The definition of $f$ being continuous at a point $c$ is that for all $\epsilon > 0$ there is a $\delta >0$ such that $\lvert x - c < \epsilon \Rightarrow \lvert f(x) - f(c) \lvert < \epsilon$.

Now let $c\in [a,b]$ be such that $f(c) > 0$. Let $\epsilon = \frac{f(c)}{2} >0$. Then by definition there is a $\delta >0$ such that if $\lvert x - c\lvert < \delta$ then $\lvert f(x) - f(c) \lvert < \epsilon$. So if $\lvert x - c\lvert < \delta$ then $$\begin{align} &\lvert f(x) - f(c) \lvert < \epsilon \Rightarrow \\ &\lvert f(x) - f(c) \lvert < \frac{f(c)}{2} \Rightarrow \\ &-\frac{f(c)}{2} < f(x) - f(c) < \frac{f(c)}{2} \Rightarrow \\ & 0< \frac{f(c)}{2} < f(x). \end{align} $$

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From the definition of continuity you have that $f$ is continuous at $x$ if for every $\varepsilon>0$ there exists $\delta>0$ such that for every $y \in (x-\delta,x+\delta)$ you have $|f(x)-f(y)|<\varepsilon$.

This means that you can choose a small neighborhood of $x$ such that $f$ is as close as you want to $f(x)$ on that neighborhood.

In your case, you know that $f(x)>0$. Then you can pick a neighborhood such that the distance from $f(x)$ to $f(y)$ is smaller than $f(x)/2$. This prevents $f(y)$ to be negative on that neighborhood.

Of course you can write all this down using epsilon and deltas, and using the triangle inequality. You should try that.

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Ok, I am going to mix and match formality and informality, here. Because I can. We are told from the outset that $f$ is continuous. In particular, this means it is continuous at $c$. What is one formal way of saying $f$ is continuous at $c$? We say that $\lim\limits_{x \to c} f(x) = f(c)$.

I am going to call $f(c)$, $u$. Why? I like that letter (Don't you?). We are also told that $f(c) = u > 0$, in short: it is positive. Now, when we say that $f$ is continuous at $c$ (that limit business), what we mean is that if $x$ is near $c$, $f(x)$ is near $u$. So surely if we pick some $x$ close enough to $c$, we ought to have $f(x)$ close enough to $u$ that it is still positive (even if it's just a little bit less than $u$, values larger than $u$ will be, of course, even more positive).

How close is "close enough"? Here the formal definition of a limit comes to our rescue. When we say $\lim\limits_{x \to c} f(x) = f(c) = u$, what we mean (formally) is given any positive number (even an itty bitty one), which we call $\epsilon$ (apparently, Greek letters are way cooler than English ones), we can find some positive number $\delta$ so that whenever $|x - c| < \delta$, then $|f(x) - u| < \epsilon$. Since we already know $f$ is continuous, we get "$\delta$" for free, anytime we pick an "$\epsilon$". So in this case, "close enough" means "within $\delta$".

Now, all we need to do, is choose $\epsilon$. Following the steps of the great set-theorist Indiana Jones, we shall "choose wisely". Our choice? We shall use $\epsilon = \frac u 2$. Our reasons are we can rest assured that:

  1. $u/2$ is just as positive as $u$ is, and

  2. It is smaller than $u$. Both of these will be important.

Now from the limit definition, and the continuity of $f$, we get our fat, grubby hands on a $\delta$ that guarantees that as long as $|x - c| < \delta$, we can be confident that $|f(x) - u| < \frac u 2$. However, what we actually want to do is prove $f$ is positive on such a range of $x$'s. So the absolute value signs aren't fair: that automatically positifies everything. So we take it away: recall that if $t < 0 < s$, then $|t| = -t > 0 > -s$. This lets us re-write the inequality $|f(x) - u| < \frac u 2$ as the double inequality:

$$-\frac u 2 < f(x) - u < \frac u 2$$

where what we have is honest-to-midwestern-goodness real numbers, no more hiding behind those absolute value signs, no sir. Now we can add $u$ to all three terms, without fear of messing up the inequalities:

$$-\frac u 2 + u < f(x) - u + u < \frac u 2 + u \iff\frac u 2 < f(x) < 3\frac u 2$$

The inequality on the right-hand isn't very interesting to us, knowing that $f(x) < 3u/2$ doesn't reveal anything about whether or not $f(x)$ is positive. But the one on the left is pure gold: it tells us that on the entire real interval $(x-\delta,x+\delta)$, that $u/2 < f(x)$. And $u/2$ is positive, so $f(x)$ is even "more positive". And on that happy note, I bid you adieu.

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Why do everybody insists on setting $\epsilon = \frac {f(c)}{2}$ if $\epsilon = f(c)$ suffices? –  Pedro Tamaroff Apr 18 '12 at 20:26
    
Any value between $0$ and $f(c)$ would do. Also, the correct word is "does" when you use a plural pronoun such as "everybody". As to why...? For clarity's sake. $f(c)/2$ is clearly less than $f(c)$, so we know that $f(x)$ stays "well above" $0$. –  David Wheeler Apr 25 '12 at 5:55
    
Bah. I think that there is no doubt that $0<f(x)<2f(c)$ when $f(c)$ is chosen. I'm guessing no-one will doubt $f(x)$ will stay away from $0$. –  Pedro Tamaroff Apr 25 '12 at 9:30
    
When doing an "epsilon-delta" proof, the point is to find a delta that works, not "the best possible delta". In this case, since we get delta for free, we need to find "an epsilon that works". No claim is made that this is "the only epsilon that works", or that it is even maximal in some sense. –  David Wheeler May 14 '12 at 15:42
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I've already given one Function Monkey-related answer today, so why not another?

Suppose we say $f(x_0) = y_0$; that is, when graph-plotting Function Monkey $f$ happens upon the coconut at $x$-value exactly $x_0$, he throws that coconut to height exactly $y_0$.

Now, in general, the Function Monkey's arm may get quite a workout, throwing coconuts high and low and all over. "Continuity of $f$" (at $x=x_0$) is the property that, when the Function Monkey is "sufficiently close" to $x_0$, he'll throw coconuts that land "arbitrarily close" to height $y_0$. Whatever that means.

Here's what that means:

When (and only when, so "whenn"?) the function is continuous ... If you give me an allowable (in)accuracy in coconut heights --if you will, an acceptable "$\epsilon$"rror (no matter how small) in computing $y_0$-- then I can construct a leash for the Function Monkey that keeps him from wandering too far away --restricting his "$\delta$"istance from $x_0$-- in such a way that every coconut the leashed Function Monkey can reach will be flung with the desired accuracy. Your flexibility to demand any level of accuracy (apart from dead-on bull's-eyes every time, because that's just too much to ask from a Monkey), and my ability to meet every such demand with an appropriately-sized leash (that allows at least some movement, because pinning a Monkey to a single point would be cruel), means that, whatever wild behavior the Function Monkey may exhibit "globally", his behavior "locally" is quite tame: the graph must "pass through" $(x_0,y_0)$ in a nice --continuous-- way.

To reiterate: For every level of accuracy you choose, I can find a satisfactory leash-length to guarantee that accuracy from the Function Monkey. In the parlance, "For every $\epsilon$, there is a $\delta$!"

In the formal definition of continuity, we specify that $\epsilon > 0$ (to allow some wiggle room $y$-wise). We also specify $\delta > 0$; I say that's because pinning the Function Monkey down is cruel, but the real reason (at least, a reason with a much smaller imaginary component) is that it prevents the smart-aleck choice of $\delta=0$ all the time. (Of course a Function Monkey standing right at $x_0$ will throw his coconut "within any $\epsilon$" of $y_0$ ... because the $x_0$ coconut lands on $y_0$! DUH! That's not helpful!) Since we're interested in showing there are coconuts "near enough" to $x_0$ that land "near" $y_0$, we have to let the Function Monkey wander just a little. (Oh, by the way: the definition's stuff about $|f(x)-y_0|<\epsilon$ and $|x-x_0|<\delta$ is just the formal way of saying "$f(x)$ is within $\epsilon$ of $y_0$" and "$x$ is within $\delta$ of $x_0$". But you knew that.)

Be that as it may ... As others have addressed, your specific problem is one in which the "$c$"oconut has been lobbed some height strictly above the $x$-axis. The word "strictly" is important here, because it means there's some (perhaps small) vertical distance between the "$c$"oconut and the ground; its also means that other coconuts can be not-quite-so-high yet still be above ground. Specifically, if the "$c$"oconut is at height $h$, then any coconut "within $h$" (vertically) of the "$c$"oconut will be above ground. This exactly describes an allowable "$\epsilon$"rror in height ---namely, $\epsilon = h$ (or any positive proper fraction thereof)--- that nevertheless guarantees above-ground-ness; consequently, supposing a continuous function, there must exist some allowable "$\delta$"istance the Function Monkey can wander that guarantees coconut-lobbing $\epsilon$-close to the "$c$"oconut, and that, in turn, guarantees above-ground-ness. As desired.

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I'm afraid you will suffer the TL;DR phenomenon, bud. –  Pedro Tamaroff Apr 18 '12 at 23:29
    
@Peter: Someone will read it. That's good enough for me. –  Blue Apr 18 '12 at 23:32
    
I'm not saying I won't either , but sometimes brevity is a virtue when explaining. Also, I think it is better to leave monkeys alone when dealing with Mathematical Analysis. I admire your enthusiasm, but I personally don't think contuinuity should be posed this way. Again, this is only my opinion. –  Pedro Tamaroff Apr 18 '12 at 23:40
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