Is $x+\frac{d}{dx}$ differential operator diagonalizable? I would like to prove or disprove the following statement:
Let $T=x+\frac{d}{dx}$ be a differential operator. Then $T$ is diagonalizable under some initial/boundary condition.
Here is my attemp to find the eigenvalues and eigenfunctions of $T$:
$$Tf=\lambda f$$
$$xf+\frac{df}{dx}=\lambda f$$
$$\frac{df}{dx}=(\lambda - x)f$$
$$\frac{df}{f}=(\lambda-x)dx$$
By integrating on both sides:$$Lnf=(\lambda x -\frac{1}{2}x^{2}+ C_{0} )$$
$$f=e^{(\lambda x -\frac{1}{2}x^{2}+C_{0} )}$$
$$f=Ce^{(\lambda x -\frac{1}{2}x^{2})}$$
I do not know how to proceed from here. For instance, if I assume $f(0)=f(1)$, I obtain $\lambda = \frac{1}{2}$ and $f(x)=Ce^{\frac{1}{2}(x - x^{2})}$. Does it say anything about $T$ being diagonalizable or not?
Update: what I mean by diagonalizable is that there exits an integral/differential operator $P$ such that T = $P^{-1}DP$ ($D$ is a diagonal operator). As an example see this .
Edit: The domain of $T$ is the set of all differentiable functions defined on $[0,1]$ with $f(0) \neq 0$.
 A: If we were dealing with linear algebra and finite matrices, this would be more clear. But we are not here. In particular, you haven't been clear by what you mean when you say "linear operator".
If by linear operator you mean an operator that sends polynomials to polynomials (which your operator does) and if by diagonalizable you mean that your operator possess an eigenvector of degree $n$ for each $n$, then it is easy to solve your problem. The answer is NO, $x+D$ is not diagolanizable. This is because there is no constant eigenvector. You can check, for any $c\not=0$, $T[c]=cx$ is also not a constant. In fact, there are no polynomial egienvectors. Another way to say this, is that there is no way to write the constant $1$ as a linear combination of vectors in the range of the linear operator, that is, there does not exist a basis of polynomials, $B_n(x)$, $\deg(B_n(x))=n$, $B_0\not=0$, and real numbers, $a_n$, such that $$1=a_0T[B_0(x)]+a_1T[B_1(x)]+\cdots$$
Again, you haven't been clear on what you are allowing for diagonalization. Or what the restrictions of your domain and range are for your operator. For example, you found the collection of all possible eigenvectors. Cool, tell me how to write $p(x)=x$ as a linear combinations of those eigenvectors. Moreover, are we regarding $p(x)$ as of infinite degree with respect to your choice of basis? Because, that isn't made clear by your statements. What are your restrictions? All possible differentiable functions? What is your domain, the uniform limit closure of your newly discovered eigenvectors? BTW, there do exist differnentiable functions with no second derivative anywhere; meaning while your domain may be differentiable functions, your range would not be.
I think you get the idea. Normally linear operators are defined on polynomials and convergently extended. If you instead want to talk about operators diagonalized by a non-polynomial basis, then you need to tell us the restrictions of that. If you simply leave it arbitrary, then it is anyone's guess what your domain and range are. It could very well be the case that your operator is not defined on polynomials at all if you choose some transcendental basis by which no polynomial can be successfully written as a summation of.
