1. ## subspaces

determine whether w={f:f(x)greater than or equal to 0} is a subspace of C, the set is continuous functions on the real numbers

2. If f is in w, think about the function -f.

3. In other words,is $\displaystyle w$ closed under scalar multiplication ?

4. Originally Posted by Raoh
In other words,is $\displaystyle w$ closed under scalar multiplication ?

No, it isn't. Look at DrSteve's post. For example choose $\displaystyle f(x)=1$ for all $\displaystyle x\in \mathbb{R}$ then, $\displaystyle f\in W$ however $\displaystyle [(-1)f](x)=(-1)f(x)=(-1)\cdot 1=-1$ . That is, $\displaystyle (-1)f\not\in W$ .

Fernando Revilla

5. Originally Posted by FernandoRevilla
No, it isn't. Look at DrSteve's post. For example choose $\displaystyle f(x)=1$ for all $\displaystyle x\in \mathbb{R}$ then, $\displaystyle f\in W$ however $\displaystyle [(-1)f](x)=(-1)f(x)=(-1)\cdot 1=-1$ . That is, $\displaystyle (-1)f\not\in W$ .

Fernando Revilla
Just my hunch, but I think Raoh understands the problem, and is trying to get the OP'er to ask the question in post #3. Could be off on that, though.

6. My question was meant to help the OP'er.

7. Originally Posted by Raoh
My question was meant to help the OP'er.

Sorry, by error I thought your post was from the original poster.

Fernando Revilla

8. Wow - it's not too hard to get thanked around here - very polite group!