# Supremum of function

1. Jan 26, 2012

### kingwinner

Let {fi}i E I be a family of real-valued functions Rn->R.
Define a function
f(x)
=sup fi(x)
i E I

1) I'm having some trouble understanding what the sup over i E I of a function of x means? The usual "sup" that I've seen is something like
supf(x)
x E S
for some set S.
But they instead have i E I there which confuses me.

2) Is the following true?
sup [c * fi(x)]
i E I

= c *sup [fi(x)]
----i E I
In other words, can we pull out a constant out of the sup? If so, how can we rigorously prove it?

Any help is appreciated!

2. Jan 26, 2012

### chiro

Hey kingwinner.

I'm pretty sure in this context E means (an element of). Basically I is a set and i is talking about referencing an element of that set. In this context, you have a collection of functions. Usually we denote things like this as a collection of whole numbers but we generalize the notation by using sets.

As far as the second question goes, the answer should be yes but only because you are doing a linear transformation. Also the other thing is that c needs to be a positive real number (> 0) otherwise you can't do this. Think about what happens when everything is multiplied by a negative number or when everything is multiplied by zero.

If you did not do a simple linear transformation like above (in terms of multiplying and adding constants or constant functions) then this doesn't need to hold. Consider if you had functions all with negative values less than one and squaring the function. What do you think would happen to the supremum?

Sometimes its helpful to draw a few diagrams.

3. Jan 26, 2012

### kingwinner

Yes, for i E I, the E means "an element of".

1) OK, so
f(x)
=sup fi(x)
i E I
means that for every fixed x, we take the sup over i E I.

2) Suppose c is a constant >0. How can we rigorously prove from the definition of sup that
c*sup [fi(x)]
i E I

=sup [c * fi(x)] ?
i E I
It's still not clear to me...

Thanks for any help!

Last edited: Jan 26, 2012
4. Jan 26, 2012

### micromass

Well, to rigorously prove that, you need to prove two things:

1) For all i holds that $cf_i(x)\leq c\cdot\sup_{i\in I}{f_i(x)}$.

2) If for all i holds that $cf_i(x)\leq M$, then $c\cdot\sup_{i\in I}{f_i(x)}\leq M$.

5. Jan 26, 2012

### kingwinner

But this only shows that

=sup [c * fi(x)] ?
i E I
≤ c*sup [fi(x)]
----i E I

6. Jan 26, 2012

### micromass

No, this shows equality. (1) shows that $\sup_{i\in I}{cf_i(x)}\leq c\sup_{i\in I}{f_i(x)}$.

7. Jan 26, 2012

### kingwinner

1)
This implies
sup [c * fi(x)]
i E I
≤ c*sup [fi(x)]
----i E I
This proves one direction.

But I don't follow your second part. What is M equal to? And why would this imply c sup f ≤ sup c f ?

Thanks.

8. Jan 26, 2012

### micromass

M is a number. Recall the definition of a supremum: a supremum is the smallest upper bound. So what I did is establish in (1) that it is an upper bound. And in (2) I establish that it's the smallest upper bound. Indeed: M is an arbitrary upper bound and I prove that M is greater than $\sup{c f_i(x)}$. This shows that M is the smallest upper bound.

9. Jan 26, 2012

### kingwinner

I see.

But I have some trouble seeing why the above is true.

I can understand the following implication
If for all i holds that $cf_i(x)\leq M$, then $\cdot\sup_{i\in I}{c f_i(x)}\leq M$. But then why can we take the c out of the sup on the LHS? I think that's what we're actually trying to prove?

10. Jan 26, 2012

### micromass

Try to do it this way:

If $cf_i(x)\leq M$, then $f_i(x)\leq \frac{M}{c}$ (if c is nonzero!!). Now take the supremum of both sides.

11. Jan 26, 2012

### SteveL27

You have a collection of functions f_1, f_2, f_3, ... f_i, ...

Think of them all superimposed on the same set of coordinate axes. So you have a bunch of function graphs on the plane.

Now for a given value of x, draw the vertical line through it. It hits EACH of the functions in one point: (x, f_1(x)), (x, f_2(x)), (x, f_3(x)), ...

Now the set of real numbers f_1(x), f_2(x), f_3(x), ... may happen to have a sup. For some values of x the sup will be defined; for other values it might not be (it might be an unbounded sequence. Then you could say the sup is +infinity if you are working in the extended real numbers).

You can define a new function f by f(x) = sup{f_i(x)} as i ranges over the index set. This new function f is well-defined for any x for which the sup exists for that particular x.

That's how to think of this.

(ps) I noticed that the domain is R^n. In that case you should still visualize the x-axis as the "domain axis," to coin a phrase. But here it's not literally true. The domain is a point in n-space so we can't visualize the graph as easily. But you can still see that the value of each function is a real number -- the range is R. So the same logic as before applies. For a fixed value of the domain, the SET of values of all the f_i may have a sup; and if it does, you can define f at that point in the domain as that sup.

Last edited: Jan 26, 2012
12. Jan 27, 2012

### kingwinner

Thank you very much.