# Homework Help: Semi-Norm Proof

1. Aug 31, 2010

### iamalexalright

1. The problem statement, all variables and given/known data
Let $$C^{1}(\Re)$$ be the space of real-valued functions which have continuous first derivatives. Let

$$||f||=sup_{-5 \leq x \leq 5} |f(x)|$$

Prove that ||.|| is a semi-norm, but not a norm on the space.

2. Relevant equations
As far as my understanding goes a semi-norm is a norm that allows some non-zero vectors to give a zero norm.

Criteria for a norm:
1. ||v|| >= 0
2. ||v|| = 0 iff v = 0 (will have to prove this isnt true for our case)
3. ||cv|| = |c| ||v||
4. ||u + v|| <= ||u|| + ||v||

3. The attempt at a solution

The way I consider the norm is as follows:
We find the supremum of the set {|f(x)|: -5 <= x <= 5}

1. From the definition we know the norm will always be greater than zero (absolute value)

2. The only way for this norm to be zero is if the function is zero on the interval (maybe this isnt the case but as far as my limited knowledge can tell it is...). This seems wrong though since if thats the case then v = 0 which would lead me to believe that it is indeed a norm and not a semi norm.

3. Easy to prove, just algebra.

4. Easy analysis proof.

Where am I wrong? What haven't I considered?

2. Aug 31, 2010

### Dick

You haven't considered that the function space is defined on R. Not [-5,5].

3. Sep 2, 2010

### iamalexalright

Well, why do I care about the whole space R if I am just looking at the supremum on [-5,5] ?

4. Sep 2, 2010

### Dick

Because if v is nonzero off of [-5,5] it's not the zero vector, but it's norm is still zero.

5. Sep 2, 2010

### iamalexalright

Okay, I see your point... well does the supremum have to be in the interval ?

6. Sep 2, 2010

### iamalexalright

Or do I just need to reword my statement for part 2? ie include the fact that the function isn't necessarily 0 but it is so on the interval

7. Sep 2, 2010

### Dick

You just need to find an example of a nonzero function with continuous derivatives which is zero on [-5,5].

8. Sep 2, 2010

### iamalexalright

Maybe I'm 'thinking' too hard but I can't think of anything like that (that has continuous derivatives)

I originally thought:
f(x) = x - 5 when x>5 and 0 otherwise but it doesnt have a continuous derivative

9. Sep 2, 2010

### Dick

Think just a little harder. Yes, define the function piecewise. Just make sure the derivatives match where the pieces join. Can't you think of a nonzero function whose derivative is zero at x=5?

10. Sep 2, 2010

### iamalexalright

f(x) = x^2 (x > 5), 0 otherwise

For some reason I was thinking functions in C1 must have _constant_ derivatives >.<

11. Sep 2, 2010

### Dick

That example isn't differentiable or even continuous at x=5. Keep thinking.

12. Sep 2, 2010

### iamalexalright

okay, to be continously differentiable the derivative has to be equal to zero at x = 5

so how about x^2 - 5x

this function is 0 at x = 5 so its continuous and the derivative is continuous at this point as well so this should be good

13. Sep 2, 2010

### Dick

The derivative of 0 at x=5 is 0. The derivative of x^2-5x at x=5 is -15. The derivative isn't continuous. The function you use to define the values for x>5 has to be zero at x=5 AND it's derivative has to zero at z=5. That's all you need. There's lots of them.

14. Sep 2, 2010

### iamalexalright

Okay... I don't know why that was so difficult for me

One such one could be:

x^2 - 10x + 25

Okay, so when writing my proving this I can just write this function as an example and not have to worry about that part anymore?

15. Sep 3, 2010

### Dick

Sure. Point out it's a nonzero function with zero norm.