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Supremum norm

  1. May 4, 2010 #1
    Could anyone tell me where to find a proof of the fact that the supremumnorm is a norm?

    The supremum norm is also known as the uniform, Chebychev or the infinity norm.
     
  2. jcsd
  3. May 4, 2010 #2

    radou

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    You could start with writing down the general definition of a norm and what it must satisfy.
     
  4. May 4, 2010 #3
    I already proved three of the conditions but I am stuck on the triangle equality. Also I wanted to see how other people did it.
     
  5. May 4, 2010 #4

    radou

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    The proof of the triangle inequality is a consequence of the "standard" triangle inequality.

    Edit: and of the property of the supremum, that sup{f(x) + g(x), x in S} <= sup{f(x), x in S} + sup{g(x), x in S}, for any functions f, g.
     
  6. May 4, 2010 #5
    The property of the supremum is exactly where I got stuck. I had the feeling that the property holds, but I could't figure out why. Could you tell me?
     
  7. May 4, 2010 #6

    HallsofIvy

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    Do you mean "[itex]sup(f+ g)\le sup(f)+ sup(g)[/itex]"? That's pretty close to trivial.

    Let x be any value. Then [itex]f(x)\le sup(f)[/itex] and [itex]g(x)\le sup(g)[/itex]. Together, [itex]f(x)+ g(x)\le sup(f)+ sup(g)[/itex]. That is, sup(f)+ sup(g) is an upper bound on f(x)+ g(x) and since sup(f+ g) is the least upper bound, [itex]sup(f+ g)\le sup(f)+ sup(g)[/itex].
     
  8. May 4, 2010 #7
    Thanks a lot. Pretty stupid I didn't think of that.
     
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