How to prove $f(x)=a$ is a subspace

Question

Prove $F$ is a subspace where $\{f \in F \mid f(x)=\text{a constant for all}\; x\}$

So far I have $f(0) = a$,

let $f,g \in F$, $(f+g)(x) = a + a $

let $k \in \mathbb{R}$, $kf(x) = k a$

But I don't think that's the proper proof, what is the proper way to prove it's a subspace?

Answer 1

Let $V$ be a vector space. A set $U \subseteq V$ is a subspace of $V$ if given $u,v \in U$, $\alpha \in \mathbb{F}$, then $u+v \in U$, $\alpha u \in U $ and zero vector is in $U$.

Notice the zero function $\Theta(x) = 0$ is a constant thus it is in set $F$. Take $f,g \in F$, then $f(x) = a$ and $g(x) = b $, where $a,b$ are cosntants, but $(f+g)(x) = f(x)+g(x) = a + b$ and $a+b$ is still a constant thus $f+g$ better be in $F$. Similarly for the last condition.

Answer 2

The space of continuous functions $\mathscr{F}$ is a vector space when the vector operations as defined as

$$ (f\oplus g)(x) = f(x) + g(x) \\ (\alpha \odot f)(x) = \alpha \cdot f(x) $$

The key point here is that we are using operations in $\mathbb{R}$ (RHS) to define operations on $\mathscr{F}$ (LHS). This is a vector space because it satisfies vector spaces axioms.

Now, we know that a subset $A$ of a vector space $V$ is a subspace if

1. Addition is closed in $A$, that is $v,w\in A\implies v\oplus w \in A$.
2. Multiplication by scalar is closed in $A$, that is, $\alpha \in \mathbb{R}, v\in A \implies \alpha\odot v\in A$.
3. The origin of $V$ is contained in $A$.

aka a subspace is a subset with the inherited vector space structure

Now, we just have to check 1, 2 and 3 for the set $F$ of constant functions. Let $f(x) = a$, $g(x) = b$ be constant functions.

1. $(f\oplus g)(x) = f(x) + g(x) = a + b $ = a constant $\implies (f\oplus g) \in F$.
2. $(\alpha \odot f)(x) = \alpha \cdot f(x) = \alpha \cdot a = $ a constant $\implies (\alpha \odot f) \in F$
3. The null vector $o(x) = 0$ is in $F$, obviously.

Writing down precisely what you are doing (e.g. define who are the functions $f$ and $g$ used in your proof, etc.) will probably help you.

Answer 3

You have the set $F$ being all constant functions (where $f(x) = a$ for $a \in \mathbb{R}$). To show that it's a vector subspace, you have to show that addition and/or scalar multiplication of any function in $F$ is also in $F$ (we call this closure).

You have to show that if you take two functions in $F$ and add them, you get another function in $F$. So if $f(x) = a$ and $g(x) = b$, where $a$ and $b$ are constants, $(f+g)(x) = a+b$. And $a+b$ is a constant, so $(f+g)(x)$ is an element of $F$.

The same procedure goes for scalar multiplication. You have most of it there - you just need to explain that when you add these two functions / scale a constant function by a scalar, they result in another constant function.

And also I would point out that you say $(f+g)(x) = a+a$ which isn't completely correct. What if $f(x) = 2$ and $g(x) = 3$? What is $(f+g)(x)$?