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Normed Vector Space

  • Thread starter poobar
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Homework Statement



Consider the vector space C[a,b] of all continuous complex-valued functions f(x), x [tex]\in[/tex] [a,b]. Define a norm ||f|| sup = max{|f(x)|, x [tex]\in[/tex] [a,b]}. (Math Note: technically we want to use sup instead of max but a physicists operational definition of max is the mathematial notion of sup).

a. Show that this is a norm.
b. Show that this norm does not satisfy the parallelogram law, ||x-y|| + ||x+y|| = 2||x||[tex]^{2}[/tex] + 2||y||[tex]^{2}[/tex]. Therefore, it cannot be an inner-product norm

Homework Equations



IF ||v|| = 0 then |v> = 0
||v1 + v2|| [tex]\leq[/tex] ||v1|| + ||v2|| (triangle inequality)
||v|| [tex]\geq[/tex] 0
||av|| = |a| ||v|| if a [tex]\in[/tex] complex


The Attempt at a Solution



I really have no idea where to start on this. i tried to apply the rules of a norm, but i am very confused. please help.
 

Answers and Replies

  • #2
HallsofIvy
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Why do you have "no idea how to start". You wrote out what is required for a norm. What happens if you put your specific definition of the norm into those?

If max(|f(x)|)= 0, what can you say about f(x)?
For any given x, [itex]|f(x)+ g(x)|\le |f(x)|+ |g(x)|[/itex].
max (a g(x)), for fixed a, is equal to a(max(g(x)).

By the way, because [a, b] is a closed interval, even the mathematical definitions of "sup" and "max" are the same.
 
  • #3
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ok, that makes sense. i guess the part i am confused about is how to go about proving that the definition fits all of the rules.

also, for the part about the parallelogram law, do i have to break f(x) into real and imaginary parts in oder to prove that it doesn't hold true for the rule? i feel like that's a dumb question, but this is part that is most confusing to me. since it is a complex valued function, this is the first thing that comes to mind.
 

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