My book says that the above set is closed under scalar multiplication and addition. I can't seem to under why the given set is closed under scalar multiplication.
For instance, if I take an even function
$u=x^2$
Now check for scalar multiplication:
$cu=c(x^2)$ if $c<0$ (like $-x^2$)
Isn't closure under scalar multiplication now fail because $-x^2$ is not an even function?
Best Answer
Yes, $f(x)=-x^2$ is an even function because it satisfies the definition: $$f(-x)=-(-x)^2=-x^2=f(x)\quad\text{ for all $x$ in the domain of $f$.}$$ In the same way you can verify that the set of even functions in $C(-\infty,+\infty)$ is closed under scalar multiplication.