This seems like a very trivial problem but I am having trouble wrapping my mind around the reasoning behind the answer.
The problem is:
Let $V$ be the (real) vector space of all functions $f$ from $\mathbb R$ into $\mathbb R$. Which of the following sets of functions are subspaces of $V$?
- all $f$ such that $f(3) = 1 + f(-5)$
And the solution I found was:
Not a subspace. Let $f(x)$ be the function defined by $f(3) = 1$ and $f(x) = 0$ for all $x \ne 3$. Let $g(x)$ be the function defined by $g(-5) = 0$ and $g(x) = 1$ for all $x \ne -5$. Then both $f$ and $g$ satisfy the condition. But $$(f + g)(3) = f(3) + g(3) = 1 + 1 = 2,$$ while $$1 + (f + g)(-5) = 1 + f(-5) + g(-5) = 1 + 0 + 0 = 1.$$ Since $1 \ne 2$, $f + g$ does not satisfy the condition.
The initial condition is very specific as to the input, you have $f(3) = 1 + f(-5)$, how can he assume that $f(x) = 0$ for all $x \ne 3$ or that $g(x) = 1$ for all $x \ne -5$? Wouldn't that be outside of the set and thus invalid for a proof?
Isn't this problem the equivalent of asking, is $g = g$ a subspace of $V$?
You define $f(3)=1$ and $f(x)=0$ for any other $x$. Hence $f(-5)=0$. And then $1=f(3)=1+0=1+f(-5)$, so it's fine.
Anyway, there is a much easier proof this set is not a subspace. If it was a subspace then the zero function would be there. But it isn't there, because if $f$ is the zero function then $f(3)=0$ but $1+f(-5)=1+0=1$.