# Supremum norm

1. May 4, 2010

### e^ipi=-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. May 4, 2010

You could start with writing down the general definition of a norm and what it must satisfy.

3. May 4, 2010

### e^ipi=-1

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.

4. May 4, 2010

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.

5. May 4, 2010

### e^ipi=-1

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?

6. May 4, 2010

### HallsofIvy

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, $sup(f+ g)\le sup(f)+ sup(g)$.

7. May 4, 2010

### e^ipi=-1

Thanks a lot. Pretty stupid I didn't think of that.