eigenvectors of continuous functions Given a continuous function $f$ on $\mathbb{R}^n$, is there a general notion of what it means for $f$ to have an eigenvalue? I would suppose that would mean there is an $\vec{x}$ and scalar $\lambda$ such that $f(x)=\lambda\cdot x$. Is this meaningful in any way? Is there a framework in which such a notion is useful?
 A: Focusing on the eigenvectors is to some extent missing the point. What is important is not eigenvectors but eigenspaces, and the reason we care about eigenspaces is that they are subspaces on which a linear transformation behaves particularly simply. (And the reason we care about eigenvectors is that they describe eigenspaces.) 
Nonlinear transformations are substantially more complicated than linear transformations. Once we know what a linear transformation does to a collection of vectors, we know what it does to every vector that is a linear combination of those vectors. Knowing what a nonlinear trasnformation does to a collection of vectors is substantially less information, and in particular knowing what a nonlinear transformation does to a particular point is hardly any information at all. 
(One exception is that it is sometimes specifically valuable to look at fixed points of a function, that is, a point satisfying $f(x) = x$. If $f$ is chosen carefully, then fixed points often have special meaning, and because of that there are a large number of fixed point theorems in mathematics.) 
A: I'll expand on my comment concerning Carleman-matrices, since your question is an open one, and there has not been some other definitive answer except that of Quiaochu. So the following is only to be understood as a possible interpretation, and I shall retract it, if a different, but definite answer, pops up. 
1) Consider some analytic function without constant term, say $f(x) = ax + bx^2 + cx^3 +  \ldots$ 
Then there is a (infinite sized) Carleman-matrix $A$ which has triangular shape. We would use this for the following notation.     
(excourse) Define the formal power series as the dot-product of the vector $A_1 = [0,a,b,c,d, \ldots] $ containing the coefficients of the function and the second vector, containing the consecutive powers of $x$, as $ V(x) = [1,x,x^2,x^3,...] $ such that the formal notation becomes $ f(x) = V(x) \cdot A_1 $ .
Now consider the vector $A_2$ of coefficients, which define the formal expression $ f(x)^2 = V(x) \cdot A_2 $ , then that vector $A_0 = [1,0,0,\ldots ]$, which defines the formal expression $ f(x)^0 = 1 = V(x) \cdot A_0$ and the set of all other vectors $A_k$ defining the formal expressions for all other consecutive powers of $f(x)^k = V(x) \cdot  A_k$ .
Then assume the "Carleman-matrix" $A$ as the concatenation of the vectors $A_k$ which allows then the notation $ V(x) \cdot A = [1,f(x),f(x)^2,f(x)^3, \ldots] $ or simpler $$ V(x) \cdot A = V(f(x)) \tag 1 $$     
Then the triangular $A$ can be diagonalized into the triangular matrices $M$ and $M^{-1} $ (let's denote it for ease of notation as $W = M^{-1}$ and the diagonal matrix $D$ which contains the consecutive powers of $a$. From the latter observation we may then sensically formulate, that $a$ is the eigenvalue of the function $f(x)$ - and the consecutive powers of $a$ such that $$ A = M \cdot \ ^dV(a) \cdot W  \tag 2$$ (where $ \ ^dV(\cdot)$ means the diagonal arrangement of a vector).    
The notion of an eigenvalue relates then to the powers of $A$ and this means iteration of functions, because $$ V(f(x)) = V(x) \cdot A \\ V(f(f(x))) = V(x) \cdot A^2 \\ \cdots \\
V(f^{\circ h}(x))) = V(x) \cdot A^h  \\ \text{ and } \\ A^h = M \cdot \ ^dV(a^h) \cdot W  \tag 3 $$ and thus relates rather to the iterations of $f(x)$ than to the derivatives (although in the given case, where $f(x)$ has no constant term and $A$ is thus triangular $a = f'(0)$ , too).
In addition, the vector of the socond column in $M$, say $M_1$, can then be seen as "the" eigenvector of the function. Let us for easeness of notation write for the h'th iterate of the function $f(x)$ beginning at $x$ : $$x_h = f^{\circ h} (x)$$ then
$$  V(x_h) \cdot M_1 = c \cdot \sigma(x_h)  = c \cdot a^h \cdot \sigma(x)  \tag 4 $$ 
(with some constant c and the function $ \sigma(x) $ called "Schroeder-function" after it's inventor, Ernst Schroeder.        
2) If the function has a constant term, $f(x) = K +ax + bx^2+ \ldots $ then the relation of its first derivative and the "eigenvalue" is mediated by a function's fixpoint, say $t$, because then the function $g(x) = f(x+t)-t = a^* x + b^* x^2+c^* x^3+\ldots $ has now no constant term and we've found another eigenvalue $a^*$ for $g(x)$ in the way as shown above. Because iterations of $f(x)$ and $g(x)$ are still simply related by $f^{\circ h}(x) = g^{\circ h}(x-t) +t $ we can now meaningfully interpret that $a^*$ as an eigenvalue of $f(x)$ (in the matrix-notation the recentering around the fixpoint can be expressed by a similarity-transformation, which for finite matrices does not affect the eigenvalues).     

Well, read all this remembering we are talking about formal power series only, and did not consider matters of convergence here. For the case 2) (where the Carleman-matrix for $f(x)$ is not triangular) and the recentered function $g(x)$ the convergence issue must be discussed even for the formal power series itself because its coefficients are now results of infinite summations. So this all above should only give an intuition how the concept of eigenvalues and eigenvectors could be meaningfully be introduced for (analytic) functions.
In the wikipedia article linked to below there are references to the Schroeder-function (and the original article of E. Schroeder) as well as to further articles, which deal with the Eigen-decomposition in the triangular and the square cases of $A$. A printed reference for the triangular case could be found in the "Advanced combinatorics" of L. Comtet written in the 1940'th if I recall correctly.
A: This does not directly answer your question, and I'm not sure if it's rigorous but:
In the same way that $Ax=\lambda x$ turns a linear transformation $A$ of a vector $x$ in a vector space $\mathbb R^n$, into a scalar operation on that vector space, you could have a linear operator on a function space (which is an example of a vector space) that is equal to a scalar operation on that function space. 
For example, the derivative operator $\frac d {dx}$ when applied to a function $f(x)$ is a linear operator. Hence for a particular function $f^*$ to be an eigenvector for the derivative operator, we would need $\frac d {dx}f^*(x)=\lambda\cdot f^*(x)$. 
That means that the set of functions $e^{\lambda_i x}$ are eigenvectors of the derivative operator $\frac d {dx}$ with eigenvalues $\lambda_i$.
Similarly, $e^{\pm\sqrt{\lambda_i}x}$ are eigenvectors of the second derivative operator $\frac {d^2}{dx^2}$.
(p.s. perhaps someone with more experience than myself can check that this is correct).
