# The Set $\{ a + b \cos(x) + c \cos(2x) \mid a, b, c \in \mathbb{R} \}$ is a Subspace in $C(\R)$

## Problem 661

Let $C(\mathbb{R})$ be the vector space of real-valued functions on $\mathbb{R}$.

Consider the set of functions $W = \{ f(x) = a + b \cos(x) + c \cos(2x) \mid a, b, c \in \mathbb{R} \}$.

Prove that $W$ is a vector subspace of $C(\mathbb{R})$.

## Proof.

We verify the subspace criteria: the zero vector of $C(\R)$ is in $W$, and $W$ is closed under addition and scalar multiplication.

First, the zero element of $C(\mathbb{R})$ is the zero function $\mathbf{0}$ defined by $\mathbf{0}(x) = 0$. This element lies in $W$, as $\mathbf{0}(x) = 0 + 0 \cos(x) + 0 \cos(2x)$.

Now suppose $f_1(x), f_2(x) \in W$, say $f_1(x) = a_1 + b_1 \cos(x) + c_1 \cos(2x)$ and $f_2(x) = a_2 + b_2 \cos(x) + c_2 \cos(2x)$. Then
$f_1(x) + f_2(x) = (a_1 + a_2) + (b_1 + b_2) \cos(x) + ( c_1 + c_2) \cos(2x)$ and so $f_1(x) + f_2(x) \in W$.

Finally, for any scalar $d \in \mathbb{R}$, we have
$d f_1(x) = (a_1 d) + (b_1 d) \cos(x) + (c_1 d) \cos(2x),$ and so $d f_1(x) \in W$ as well.

This proves that $W$ is a subspace of $W$.

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