# Squeeze Theorem: What does it mean for a function to be less than or greater than a another function

1. Oct 5, 2014

### FredericChopin

The Squeeze Theorem states that if three functions $f(x)$, $g(x)$, and $h(x)$ exist, and $f(x) \leq g(x) \leq h(x)$, and if $\lim_{x \rightarrow a} f(x) = L$ and $\lim_{x \rightarrow a} h(x) = L$, then $\lim_{x \rightarrow a} g(x)$ must exist and its value must be L.

Here is something I don't understand about the Squeeze Theorem: what does it mean for a function to be less than or greater than a another function? It doesn't make sense to me because depending on the input values of the function, sometimes the value of the function can be greater than the value of a different function, and sometimes the value of the function to be less than the value of the other function, and sometimes the value of the two functions could be equal. How does one know if a function is less than or greater than another function? Can you determine it mathematically, or do you have to look at a table or a graph?

Thank you.

2. Oct 5, 2014

### Char. Limit

The usual idea is that a function, say, g(x), is greater than a function f(x) at every value in a given interval containing the point a that you're finding the limit of. After all, outside that interval, it doesn't really matter, so as long as you can prove that, say, g(x) > f(x) at every value in that interval (containing a), you're good.

3. Oct 5, 2014

### Fredrik

Staff Emeritus
I would be careful with the notation and terminology in discussions like this. f(x) isn't a function. It's an element of the codomain of the function f, so in this case (where f,g,h are real-valued functions) it's a real number. f(x) is a "function of x" in the sense that its value is completely determined by the value of x, but that doesn't mean that it's a function.

If f and g are two real-valued functions with the same domain X, then we say that $f\leq g$ if $f(x)\leq g(x)$ for all $x\in X$.

That's the definition. What this says about the graphs is that if $f\leq g$, then the graph of f never goes above the graph of g.

Note that you can't skip the "for all x" part of the statement in the definition. The inequality $f(x)\leq g(x)$ is just a part of the full statement that's useless on its own.

You asked how one proves that $f\leq g$. Typically you start by saying "Let $x\in X$ be arbitrary". Then you show that $f(x)\leq g(x)$. (If you succeed, you will have proved the statement "for all x in X, we have $f(x)\leq g(x)$", which is by definition equivalent to the statement $f\leq g$). How to do that depends on f and g of course.

Last edited: Oct 5, 2014
4. Oct 6, 2014

### FredericChopin

How so? How is using one element from the domain proving that a function is greater than another function for all elements of the domain? Could you show me an example of how to do that?

5. Oct 6, 2014

### Fredrik

Staff Emeritus
Since I said that x is an arbitrary element of the domain, I'm not allowed to use any properties of x in the next step, other than the ones that all elements of the domain have. So if I manage to prove that x has a property P, then we can conclude that every element of the domain has the property P.

Theorem: For all $x\in\mathbb R$, we have $x^2\geq 2x-1$.
Proof: Let $x\in\mathbb R$ be arbitrary. We have $0\leq(x-1)^2=x^2+1-2x$. This implies that $x^2\geq 2x-1$.

Corollary: Let f be the function defined by $f(x)=x^2$ for all $x\in\mathbb R$. Let g be the function defined by $g(x)=2x-1$ for all $x\in\mathbb R$. We have $g\leq f$.
Proof: For all $x\in\mathbb R$, we have $g(x)=2x-1\leq x^2=f(x)$. This means that $g\leq f$.

Note that the following statements mean exactly the same thing:

1. For all x in S, x has property P.
2. Let x in S be arbitrary. x has property P.

The second version is just a way to break the statement up into several sentences. This is very useful in proofs.

6. Oct 8, 2014

### FredericChopin

I see. However...

... can you explain what you did here?

7. Oct 8, 2014

### PeroK

Here's a simpler example:

$f(x) = x, \ \ g(x) = x + 1$

$\forall x \ \ f(x) < g(x)$

Hence $f < g$

8. Oct 8, 2014

### Fredrik

Staff Emeritus
Since x is a real number and 1 is a real number, x-1 is a real number. This and the fact that the square of a real number can't be negative together imply that $0\leq(x-1)^2$. The equality $(x-1)^2=x^2+1-2x$ is a special case of $(a-b)^2=a^2+b^2-2ab$ which holds for all real numbers $a,b$. The equality sign means that the expressions $(x-1)^2$ and $x^2+1-2x$ represent the same number. So $x^2+1-2x$ represents the same number as $(x-1)^2$, which we know is non-negative. This means that $x^2+1-2x$ is non-negative, i..e that $x^2+1-2x\geq 0$. Now add $2x-1$ to both sides of this inequality, and you get $x^2\geq 2x-1$.

Edit: I made a change in the text above after it was quoted below. I had incorrectly called the formula $(a-b)^2=a^2+b^2-2ab$ "the conjugate rule". I don't remember what it's called in English at the moment, but it's not that. In Swedish, the formula $(a+b)(a-b)=a^2-b^2$ has a name similar to "the conjugate rule", but that's not the formula I used here.

Last edited: Oct 19, 2014
9. Oct 9, 2014

### FredericChopin

Wow. Very elaborate.

So to find the necessary functions to use for the Squeeze Theorem, it is a matter of algebraic manipulation and the properties of numbers, like you did, right?

10. Oct 9, 2014

### Fredrik

Staff Emeritus
Yes, you can say that. Although, you can say that about almost every problem that involves real numbers.

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