Question

If V is the vector space from R to R, is the set of polynomial functions a subset of V? (I know I have to show the set of polynomials isn't empty, and is closed under addition and scalar multiplication. But with it being generic, with no functions given, I am a little stuck)

Answer 1

Here I will assume that
`V` is the vector space of functions
`f:\mathbb{R}\rightarrow\mathbb{R}`. If
`V` is the space of continuous functions, there will be no changes in the proof, because polynomial functions are also continuous.

Let us write
`P` for the set of polynomials with real coefficients. As we want to prove that
`P` is a subspace of
`V` we need to check three conditions:

First:
`P` is a non empty set, in particular
`0\in P`. This is not difficult, because the function identically 0 is a polynomial. Then,
`P` is not empty and contains the 0 vector of
`V`.

Second: The addition of two elements of
`P` stays in
`P`. Consider two arbitrary polynomials
`p` and
`q`. Then,

`p(x) = a_0+a_1x+\cdots+a_nx^n`

`q(x) = b_0+b_1x+\cdots+b_mx^m`.

This means that
`p` is a polynomial of degree
`n` with coefficients
`a_k`, and
`q` is a polynomial of degree
`m` with coefficients
`b_k`. With out lose of generality assume that
`m\geq n`.

Now, notice that

`p(x)+q(x) = (a_0+b_0) + (a_1+b_1)x + \cdots+(a_n+b_n)x^n +b_(n+1)x^(n+1) +\cdots +b_mx^m`

which is a polynomial too.

Hence,
`p+q\in P`.

Third: The multiplication by a scalar stays in
`P`.

Consider an arbitrary polynomial

`p(x) = a_0+a_1x+\cdots+a_nx^n`

and a real number
`\alpha`.

Then,

`\alpha p(x) = (\alpha a_0)+(\alpha a_1)x+\cdots+(\alpha a_n)x^n`

which is a polynomial.

Hence,
`\alpha p\in P`.

Therefore,
`P` is a subspace of the vector space
`V`.