Let $V = C^{\infty}(\mathbb{R},\mathbb{R})$ and $T \in L(V)$ defined by $(T(f))(t)=tf(t)$. Prove that $T$ has no eigenvalues. 
Let $V = C^{\infty}(\mathbb{R},\mathbb{R}) = \{ f : \mathbb{R} \rightarrow \mathbb{R} \mid f$ in infinitely many times differentiable $\} $ be a vector space. Let $T \in L(V)$ defined by $(T(f))(t)=tf(t)$. Prove that $T$ has no eigenvalues.

My attempt: 

Let $T(f) = \lambda f $ for $\lambda \in \mathbb{R} $, then $(T(f))(t) = \lambda f(t) $ for all $t \in \mathbb{R} $. Which implies that $tf(t)=\lambda f(t)$ and for non zero function $f$, this gives us $t=\lambda$. But since it should be true for all $t \in \mathbb{R}$, $T$ has no eigen values. 

Is this proof correct? Are there any other ways to solve this problem?   
 A: You may want to be more clear and explicit: since $\;f\neq0\;$ (as we assume it is an engenvector), then there exists $\;t_0\in\Bbb R\;$  such that $\;f(t_0)\neq0\implies\;$ there exists a neighborhood $\;I_0\;$ of zero where $\;f\;$ keeps the same sign as $\;f(t_0)\;$ (from continuity), so:
$$t_0f(t_0)=\lambda f(t_0)\implies t_0=\lambda$$
but the above exactly will yield $\;\lambda =t\;$ for any $\;t\in I_0\;$, and this is a contradiction.
A: I think your reasoning is very correct and explicit enough actually. I also think this result would still hold if the functions are not continuous what so ever (as a long as they from a vector space) and as long as it is not just the zero function. Not correct, as pointed out below!
Just for entertainment also notice that the vector space that just contains the zero vector is a vector space, but any linear map defined on it, can not have by definition any eigenvalues.
The main point here is, that as long as the vector space is finite dimensional a lot of nice properties hold for linear maps that are not true at all for infinite dimensional vector spaces and linear maps. One example has to do with the injectivity and surjectivity of a linear map. So any linear map on a finite dimensional vector space is injective if and only if it is surjective. And this fact enters a lot in the discussion of eigenvalues on finite dimensional vector spaces. 
