Attempting to prove the Intermediate Value Theorem

[Eclair_de_XII]

Homework Statement

Define a continuous function $f:[a,b]\longrightarrow \mathbb{R}$ such that $f(b)>0$ and $f(a)<0$. Prove that there is a real number $c\in (a,b)$ such that $f(c)=0$.

Relevant Equations

A function $f$ is continuous at $a\in \textrm{dom}(f)$ if for all positive $\epsilon$, there is a $\delta>0$ such that for $x\in \textrm{dom}(f)$, if $0<|x-a|<\delta$ then $|f(x)-f(a)|<\epsilon$

Proof goes like this:
(1) Prove the existence of open intervals centered around the end-points of the domain such that the image of the points in these intervals through $f$ has the same sign as the image of the end-point through $f$. In other words, prove that there is a $\delta>0$ such that if $b>x>b-\delta$, $f(x)f(b)>0$; and similarly for $a$.

(2) Conclude there is a point $c\in (a,b)$ such that for any $\delta>0$, the image of $(c-\delta,c+\delta)$ contains both positive and negative points.

(3) By definition of continuity, there must be a real number $L$ that $f(c)$ converges to. Reason that $L$ cannot be negative or positive.

===(1)===
Suppose that there are no such intervals. Consider $b$. Choose $\epsilon\leq f(b)$. Hence, for any $\delta>0$, if $|b-x|<\delta$, then $f(x)<0$.
\begin{align*}
|f(x)-f(b)|&=&f(b)+|f(x)|\\
&\geq&f(b)\\
&\geq&\epsilon
\end{align*}
This contradicts the fact that $f$ is continuous. Hence, there must be some $\delta>0$ such that $f(x)>0$ if $b-\delta<x<b$.

===(2)===
Now choose $c=b-\delta_b$ where $\delta_b>0$ with the property that if $x<b-\delta_b$, then $f(x)\leq 0$.
Similarly, $c=a+\delta_a$ where $\delta_a>0$ with the property that if $x>a+\delta_a$, then $f(x)\geq 0$.
% Note: I feel like I should be invoking completeness here, but wasn't sure if it applies or if it does, how to invoke it.

Let $\delta_c>0$.

If $\delta_c+c>x>c=a+\delta_a$, then $f(x)\geq 0$.
If $c-\delta_c<x<c=b-\delta_b$, then $f(x)\leq 0$.

===(3)===
Suppose $f(c)>0$ and choose $\epsilon\leq f(c)$. Let $\delta>0$ and choose $x\in (c-\delta,c)$.

Then $f(x)\leq 0$ and:
\begin{align*}
|f(c)-f(x)|&=&-f(x)+f(c)\\
&=&|f(x)|+f(c)\\
&\geq&f(c)\\
&\geq&\epsilon
\end{align*}
Hence, $f$ cannot be continuous at $x=c$ if $f(c)>0$.

Now suppose $f(c)<0$ and choose $\epsilon\leq -f(c)$. Let $\delta>0$ and choose $x\in (c,c+\delta)$.

Then $f(x)\geq 0$:
\begin{align*}
|f(x)-f(c)|&=&f(x)+(-f(c))\\
&=&|f(x)|-f(c)\\
&\geq&-f(c)\\
&\geq&\epsilon
\end{align*}
Hence, $f$ cannot be continuous at $x=c$ if $f(c)<0$.

This leaves only the one possibility by the trichotomy of the real numbers.

 

[Office_Shredder]

I don't understand step 2. Why can I pick a single c which satisfies both the condition for b and the condition for a? It feels like that's the whole problem.

 

[pasmith]

Alternatively: Define $P \subset [a,b]$ such that $x \in P$ if and only if $f(y) < 0$ for all $y \in [a,x]$. Then show $f(\sup P) = 0$.

 

[Eclair_de_XII]

Why can I pick a single c which satisfies both the condition for b and the condition for a?

Define $D_b=\{\delta>0:f(b)f(x)>0\textrm{ if }x\in(b-\delta,b)\}$.

We reason that this is bounded. Suppose $\delta=|b-a|$. Then the interval $(b-|b-a|,b)=(a,b)$ contains points from the interval around $a$ such that $f(x)<0$. Hence, this choice of $\delta$ cannot belong to $D_b$. If $\delta>|b-a|$, then for $x\in(b-\delta,b)$, $f(x)$ is not even defined. Hence, $D_b$ is bounded from above by $|b-a|$. By the completeness of real numbers, there must exist a $\sup D_b$; denote it as $\delta_b$.

If we define $D_a$ similarly, then we can make a symmetric argument.

I was going to reason that $a+\delta_a=b-\delta_b$, then call it $c$.

By definition, if $a<x< a+\delta_a$, then $f(x)<0$; if $x\geq a+\delta_a$, then $f(x)\geq 0$.
By definition, if $b>x> b-\delta_b$, then $f(x)>0$; if $x\leq b-\delta_b$, then $f(x)\leq 0$.

This is what I have so far; as for how to figuratively connect the dots, I do not know.

One thing that comes to mind is just repeating step three, replacing $c$ with either $a+\delta_a$ or $b-\delta_b$.

Then show $f(\sup P)=0$.

I feel like I'd be repeating step three with this, to be honest, if I replaced $c$ in my post with $\sup P$.

 

[Office_Shredder]

Eclair, try applying your way of computing c to the function $(x-1)(x-2)(x-3)$ on the interval [0,4]. I think you will find you did not succeed at finding a single c. But you should consider what you did find, and if you actually care about those two things lining up. This basically just turns into what pasmith is suggesting.

 

[Eclair_de_XII]

try applying your way of computing c to the function $(x-1)(x-2)(x-3)$ on the interval [0,4].

Okay.

$a=0$
$b=4$

$D_0=\{\delta>0:f(x)f(0)>0\textrm{ if }x\in(0,\delta)\}$

Assert $\sup D_0=1$ and note that $f(0)=(-1)(-2)(-3)=-6<0$.

Note that if $0<x<1$, then:

$-1<x-1<0$
$-2<x-2<-1$
$-3<x-3<-2$

Hence, the function $(x-1)(x-2)(x-3)$ is a product of negative numbers if $x\in (0,1)$; the function must be negative as a result.

If $x\geq1$:

$x-1\geq 0$
$x-2\geq -1$
$x-3\geq -2$

And the last two numbers are of some concern. I could only guarantee that they would both be negative, thus guaranteeing that the function is positive, but only if $x\leq 2$. So it wouldn't be the case that $f(x)>0$ if $x\in (1-\delta,1+\delta)$ for any positive $\delta$.

But you should consider what you did find, and if you actually care about those two things lining up.

I think I found the value for $\sup P$, which is $a+\delta_a$ and the value for $\inf([a,b]\setminus P)$, which is $b-\delta_b$, where $P$ was defined by Mr. pasmith. I mean, there would be some neighborhood around these two points that would contain points whose images are positive and negative. They'd be the same if the function was zero only at one point.

 

[Office_Shredder]

Eclair, why are you evaluating the individual signs of that factored polynomial? f is a function, you can evaluate the function but shouldn't do anything else with it

I think you should try to find $\delta_b$ to see why your method doesn't work on this function.

Remember, your goal is to find a zero of f. When finding $\delta_a$, you found the number 1. I am impressed by your ability to not just evaluate f(1) and wonder about it, but I recommend you do that now :)

 

[Eclair_de_XII]

$f(1)=(1-1)(1-2)(1-3)=0(-1)(-2)=0$

 

[Office_Shredder]

Ok, so you're done for this function, right? You computed $\delta_a$, and it found you a zero, which was your goal. Is it a coincidence that this worked, or can you turn it into a proof?

 

[Eclair_de_XII]

turn it into a proof

No, it was not really a coincidence. There are a bunch of points that the function evaluates to positive values immediately after $x=1$, and the function cannot just jump from a positive number to a negative number, since otherwise it would not be continuous.

 

[Office_Shredder]

No, it was not really a coincidence. There are a bunch of points that the function evaluates to positive values immediately after $x=1$, and the function cannot just jump from a positive number to a negative number, since otherwise it would not be continuous.

What if instead the function was $(x-1)^2(x-2)$? What is $\delta_a$? Is the function positive afterwards?

 

[Eclair_de_XII]

The value of $x$ does not affect the sign of $(x-1)^2$. And $x-2$ is negative if $0<x<2$ and positive if $4>x>2$. $\delta_a$ would be $2$ in this case; easy to see if we just subtract two from these inequalities.

 

[Office_Shredder]

No, $\delta_a=1$. Please think about this carefully and let me know if you see why.

 

[Eclair_de_XII]

$D_0:=\{\delta>0:f(x)f(0)>0\textrm{ if }x\in(0,\delta)\}$

Set $\delta=2$ and choose $x=1.5$. Note that $1.5\in(0,2)$.

$f(x)=(x-1)^2(x-2)$ on the interval $[0,4]$.

$f(1.5)=0.5^2(-0.5)$.
$f(0)=-2$
$f(1.5)f(0)=0.25>0$

Hence, $2\in D_0$.

I'm not even sure that $1$ is an upper bound for $D_0$.

 

[Eclair_de_XII]

In any case, can I ask if all that I'd written up until post #4 would have been correct if $f$ was identically zero at just one point on its domain?

 

[Office_Shredder]

What is f(1)? If $\delta_a >1$, it needs to be negative the way you defined it.

 

[Eclair_de_XII]

Oh. $f(1)=0$, so $f(0)f(1)=0$, which means that if $\delta>1$, then you can choose the point $x=1$ in $(0,\delta)$ such that $f(0)f(1)$ is not positive, which would mean that $\delta\notin D_0$.

 

[Office_Shredder]

Yes.

So how does that help you? You know that $D_0$ is bounded above by $b-a$, $D_0$ is not empty, so it has a supremum. In both of the examples we looked at, the supremum had $f(c)=0$. Is that always true?

 

[Eclair_de_XII]

Whenever $x\in(a,a+\delta_a)$, $f(x)<0$. If $x\geq a+\delta_a$, then $f(x)\geq0$. Let's say that $x=a+\delta_a$. Then $f(x)=0$ or $f(x)>0$. If the latter held, then the function would fail to be continuous at that point.

Denote $L=f(a+\delta_a)$ and suppose it is positive. Set $\epsilon\leq L$ and let $\delta>0$. If $x\in (a+\delta_a-\delta,a+\delta_a)$, then $f(x)<0$. Hence,
\begin{align*}
|L-f(x)|&=&L+(-f(x))\\
&=&L+|f(x)|\\
&\geq&L\\
&\geq&\epsilon
\end{align*}

This leaves just the other possibility, I think.

 

[Office_Shredder]

No, on

Whenever $x\in(a,a+\delta_a)$, $f(x)<0$. If $x\geq a+\delta_a$, then $f(x)\geq0$.

No, in the example we saw, $f(x)<0$ for $x>a+\delta_a$. We know $f$ is continuous, and $f(x)<0$ for $x<a+\delta_a$. Use that to show $f(a+\delta_a)\leq 0$. Then suppose it's strictly less than zero- you can show that if that's true, $\delta_a$ is not actually the supremum, which is a contradiction.

 

[Eclair_de_XII]

Let $\epsilon>0$. Then there is a $\delta>0$ such that if $|x-(a+\delta_a)|<\delta$, then $|f(x)-f(a+\delta_a)|<\epsilon$. This implies that:

$f(a+\delta_a)<f(x)+\epsilon<\epsilon$
$f(a+\delta_a)-\epsilon<f(x)<0$

This implies that $f(a+\delta_a)$ cannot be greater than some arbitrarily small $\epsilon>0$. Hence, either $f(a+\delta_a)=0$, or $f(a+\delta_a)<0$. It cannot be positive for the reason in post #19.

 

[Office_Shredder]

Why is $f(x)+\epsilon < \epsilon$?

 

[Eclair_de_XII]

Because $f(x)<0$ if you choose $x<a+\delta_a$. Adding $\epsilon$ to both sides of the former inequality gives that result.

Specifically:

$|f(a+\delta_a)-f(x)|<\epsilon$
$-\epsilon<f(a+\delta_a)-f(x)<\epsilon$
$f(a+\delta_a)<\epsilon+f(x)$

 

[Office_Shredder]

But all you have is $|x-(a+\delta_a)|< \delta$. So x could be as large as $a+\delta_a+\delta$

 

[Eclair_de_XII]

You can also choose $x$ to be as small as $(a+\delta_a)-\delta$, though.

And $f(x)$ is negative if $(a+\delta_a)>x>(a+\delta_a)-\delta$, no matter how small $\delta$ is.

On the other hand, if $f(x)\geq0$ whenever $(a+\delta_a)+\delta>x>(a+\delta_a)$:

$f(x)-f(a+\delta_a)<\epsilon$
$-f(a+\delta_a)<\epsilon-f(x)\leq\epsilon$
$-f(a+\delta_a)-\epsilon<-f(x)\leq0$
$f(a+\delta_a)+\epsilon>f(x)\geq0$

This implies that if $f(a+\delta_a)<0$ cannot exceed some arbitrarily small $\epsilon$ in magnitude.

This means that $f(a+\delta_a)\geq0$. In addition, we also have another possibility for when $x$ is on the left-side of this interval: $f(a+\delta_a)\leq0$. If $f(a+\delta_a)\neq0$, then we have continuity issues as discussed previously. I think I covered this in the third step in the opening post of this topic?

Also, if $f(x)<0$ whenever $(a+\delta_a)+\delta>x>(a+\delta_a)$, this would contradict yet again the fact that $\delta_a$ is the supremum for $D_a$.

 

[Office_Shredder]

I feel like you are way too ready to split into cases about the various signs of f(x) on the other side of the zero. Your proof feels about right but it has so many extra unnecessary things thrown in it's hard to be sure.

Here's how I would do it. Notice the conservation of detail. You proved already that if $f(x)$ is not zero then there is some interval around which it's always the same sign, so there's no need to keep proving it over and over again.

If $f(a+\delta_a) > 0$, then by step 1 of the proof there is some $\delta > 0$ such that $f(a+\delta_a -\delta/2)> 0$. This contradicts the fact that $f(x)<0$ for $x<a+ \delta_a$

If $f(a+\delta_a) < 0$, then by step 1 of the proof there is some $\delta > 0$ such that if $a+\delta_a < x a+\delta_a + \delta$, $f(x)<0$. But that means if $a<x < a+\delta_a+\delta$ then $f(x)<0$, which contradicts the fact that $\delta_a$ was the supremum of the set.

Therefore $c=a+\delta_a$ has $f(c)=0$.

 

[Eclair_de_XII]

Your proof feels about right but it has so many extra unnecessary things thrown in it's hard to be sure.

Many of my professors told me pretty much the same thing when I was going to school as an undergraduate student over two years ago. They expressed that my proofs were too convoluted, and were too long, often spanning close to, if not over, a page. In any case, I appreciate all the trouble you went through for guiding me towards understanding the IVT.