Proving ||.|| is a Semi-Norm on C^1(\Re)

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SUMMARY

The discussion centers on proving that the supremum norm defined as ||f||=sup_{-5 ≤ x ≤ 5} |f(x)| is a semi-norm on the space C^{1}(\Re) of real-valued functions with continuous first derivatives, but not a norm. The key criteria for a norm are outlined, including non-negativity, zero norm implying the zero vector, homogeneity, and the triangle inequality. The participants conclude that a nonzero function can exist which is zero on the interval [-5, 5], thus demonstrating that the supremum norm does not satisfy all the conditions of a norm.

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  • Understanding of C^{1}(\Re) and continuous first derivatives
  • Knowledge of supremum norms and their properties
  • Familiarity with the definitions and criteria of norms in functional analysis
  • Ability to construct piecewise functions with continuous derivatives
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  • Study the properties of semi-norms and norms in functional analysis
  • Learn about piecewise continuous functions and their derivatives
  • Explore examples of functions in C^{1}(\Re) that demonstrate the properties of norms and semi-norms
  • Investigate the implications of nonzero functions having zero norms in various function spaces
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Mathematics students, particularly those studying functional analysis, researchers in mathematical analysis, and educators looking to deepen their understanding of norms and semi-norms in function spaces.

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Homework Statement


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.


Homework 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 isn't true for our case)
3. ||cv|| = |c| ||v||
4. ||u + v|| <= ||u|| + ||v||



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 isn't the case but as far as my limited knowledge can tell it is...). This seems wrong though since if that's 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?
 
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You haven't considered that the function space is defined on R. Not [-5,5].
 
Well, why do I care about the whole space R if I am just looking at the supremum on [-5,5] ?
 
iamalexalright said:
Well, why do I care about the whole space R if I am just looking at the supremum on [-5,5] ?

Because if v is nonzero off of [-5,5] it's not the zero vector, but it's norm is still zero.
 
Okay, I see your point... well does the supremum have to be in the interval ?
 
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
 
iamalexalright said:
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

You just need to find an example of a nonzero function with continuous derivatives which is zero on [-5,5].
 
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 doesn't have a continuous derivative
 
iamalexalright said:
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 doesn't have a continuous derivative

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

For some reason I was thinking functions in C1 must have _constant_ derivatives >.<
 
  • #11
iamalexalright said:
f(x) = x^2 (x > 5), 0 otherwise

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

That example isn't differentiable or even continuous at x=5. Keep thinking.
 
  • #12
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
iamalexalright said:
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

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

One such one could be:

x^2 - 10x + 25Okay, 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
iamalexalright said:
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?

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

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