subspace of the Vector Space of real valued functions

This is a problem from Hoffman and Kunze's Linear Algebra 2nd edition. I am trying to determine whether or not a particular subset of the set of all real valued functions is a subspace. I've done linear algebra before in college, but we primarily dealt with the real numbers, sometimes complex numbers, matrices, and really that was it. I think I"ve done the problem correctly, but some proof verification would help.

Okay so $V$ is the set of all real valued functions from $\mathbb{R}$ into $\mathbb{R}$. Take a subset $$W=\{f|f(3)=1+f(-5)\}$$ So I wracked my brain a bit and thought of a function that increases by 1 every 8 units and determined that the easy fit would be $f(x)=\frac1{8}x$. So $f\in W$ since $$f(3)=\frac{3}{8}=\frac{8-5}{8}=1+\frac{-5}{8}=1+f(-5)$$ and $W$ is nonempty. But now, take $k\in \mathbb{R}$, such as $k=3$ Now consider $3f$. $$(3f)(3)=3\frac{3}{8}=\frac{9}{8}\neq\frac{-7}{8}=\frac{8-15}{8}=1+\frac{-15}{8}=1+3\frac{-5}{8}=1+(3f)(-5)$$ So since this is condition is not met, this can not be a subspace of real valued functions.

Another question I have is this: I know there are infinitely many functions simply because I could say $f(x)=\frac{1}{8}x+c$, where $c\in\mathbb{R}$. Also, you could apply the appropriate transformation to the sine function without adjusting the amplitude and probably make it so it satisfies as the condition as well. But all of these examples are more counter examples. Is there any example where the condition is satisfied and a single function under these conditions provides the necessary and sufficient conditions to be a subspace?

• You proved that $W$ is not a subspace of $\Bbb R^{\Bbb R}$. It's not a subspace of anything because $0\not\in W$. – leo Sep 20 '14 at 2:20
• well, that would have been a much easier approach. I should have considered that. – Lalaloopsy Sep 20 '14 at 2:40