Is $f(x)=kf(-x)$ a subspace when $k$ is fixed? Let $V$ be the set of all real valued functions defined on real numbers. Fix $k \in \mathbb{R}$ and let $W_k= \{f \in V: f(x) = kf(-x) \,\forall\, x \in \mathbb{R} \}$.
I think one of the axioms fails for $W_k$ to be a subspace of $V$. If I multiply the scalar $1/k$ to $f(x)$, I get $f(-x)$ which doesn't belong to $W_k$. But the correct answer states that $W$ is a subspace. 
I can see the other two axioms being satisfied here but the scalar multiplication troubles me. $1/k$ does belong to $\mathbb{R}$ and in the function defined $k$ is fixed. 
 A: If $f(x)=kf(-x)$ then $(\frac  1 k f(x))= k (\frac  1 k f(-x))$ so $(\frac  1 k f(x))$ so also in this set. 
More generally if $f$ and $g$ are in $W$, $a,b $ are scalars and $h=af+bg$ then $h(x)=a(k(f(-x))+b(kg(-x))=k[af(-x)+bg(-x))]=kh(-x)$ so $h \in W$. 
A: V = set of all people in your street and W = set of all people in your street whose name starts with a. Now, if you want to show your neighbor friend is in W you have to verify that his name starts with a. Because this is the condition for a person to be in W. 
Now, coming to Mathematics, a function $f$ is in $W_k$ if and only if it satisfies $f(x) = k(f(-x))$. So to show some $f$ is is $W_k$, you have to verify this condition $f(x) = k(f(-x))$. 
To verify $W_k$ is a subspace we have to verify that if $f,g \in W_k$ then $\alpha f + \beta g \in W_k$. Since $f$ and $g$ are in $W_k$ we have $$f(x) = k(f(-x))$$ and 
$$g(x) = k(g(-x)).$$
To prove $\alpha f + \beta g \in W_k$ we have to prove 
$$(\alpha f + \beta g)(x) = k(\alpha f + \beta g)(-x)$$
Let's prove LHS = RHS.
$$(\alpha f + \beta g)(x) = \alpha f(x) + \beta g(x) = \alpha (k (f(-x))) + \beta (k (g(-x))) $$
$$= k(\alpha (f(-x)) + \beta  (g(-x))) = k(\alpha f + \beta g)(-x)$$
