# Determine if L is a vector space

• Dank2

## Homework Statement

V = function space from R to R
L ={ f in V | f(1/2) > f(2) }
Determine if L is a vector space.

## The Attempt at a Solution

1. Closed under addition: Do i do addition like this let g and e in V, then g(1/2)+e(1/2) > g(2) + e(2) but the addiction of two functions is even a function ??

## Homework Statement

V = function space from R to R
L ={ f in V | f(1/2) > f(2) }
Determine if L is a vector space.

## The Attempt at a Solution

1. Closed under addition: Do i do addition like this let g and e in V, then g(1/2)+e(1/2) > g(2) + e(2) but the addiction of two functions is even a function ??

Of course it is. After all, what is a function, anyway?

Besides addition, you also need to look at multiplication by a scalar, that is, you need to look at the function c*f(x), where c is a number and f is a function.

Dank2
Of course it is. After all, what is a function, anyway?

Besides addition, you also need to look at multiplication by a scalar, that is, you need to look at the function c*f(x), where c is a number and f is a function.
So as the way i have shown above is enough for proving its closed under addition?
let a be a scaler, and f be in V then a*(f(1/2)>a*f(2) is that enough for showing it's closed under scalar multiplication ? looks too much trivial

If your claim were true, yes, that's all you have to do. You should think about the different types of values ##a## can assume.

If a = 0 inequality won't hold
Therefore it's not vector space above r?

If a = 0 inequality won't hold
Therefore it's not vector space above r?

If f(1/2) = 10 and f(2) = 2, we have f(1/2) > f(2) because 10 > 2. Do we also have -10 > -2?

If a = 0, inequality won't hold; therefore, it's not vector space above r?
Yes, that's right. You can also use what Ray pointed out.