Where can I find a proof of the supremum norm as a norm?

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The supremum norm, also known as the uniform, Chebychev, or infinity norm, is confirmed to be a norm through the proof of the triangle inequality. The discussion highlights that the triangle inequality follows from the standard triangle inequality and the property of the supremum, specifically that sup{f(x) + g(x), x in S} ≤ sup{f(x), x in S} + sup{g(x), x in S}. This property establishes that the supremum of the sum of two functions is less than or equal to the sum of their suprema, thus validating the supremum norm as a norm.

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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.
 
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You could start with writing down the general definition of a norm and what it must satisfy.
 
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.
 
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.
 
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?
 
Do you mean "sup(f+ g)\le sup(f)+ sup(g)"? That's pretty close to trivial.

Let x be any value. Then f(x)\le sup(f) and g(x)\le sup(g). Together, f(x)+ g(x)\le sup(f)+ sup(g). 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, sup(f+ g)\le sup(f)+ sup(g).
 
Thanks a lot. Pretty stupid I didn't think of that.
 

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