what is $\dim \{f\in P_n|T(f)=f\}$ Let $P_n$ be polynomials of degree $\leq2n$ on $\mathbb{R}$.Clearly $P_n$ is a finite dimensional vector space.Let
$$ 
T(f)=\int_{\mathbb{R}}e^{(x^2-y^2)\pi}\cos(2\pi yx)f(y)dy
$$
It's easy to show that $T$ maps $P_n$ to itself. 
The problem is what is $\dim \{f\in P_n|T(f)=f\}$.
My idea:
Suppose $f(x)=x^k$, we can observe that $T(x^k)=0$ when $k$ is odd. When $k$ is even, by Fourier transformation,I prove that
$$
T(f)=\int_{\mathbb{R}}e^{(x^2-y^2+2ixy)\pi}f(y)dy=\frac{1}{(2\pi i)^k}e^{x^2\pi}\frac{\mathrm{d}^k}{\mathrm{d}x^k}e^{-\pi x^2}
$$
When $k$ is even, it's easy to prove $T(x^k)$ is a polynominal with degree $k$, and its items are only even items.
And let $\{x^k\}_{k=0}^{n}$ be a basis. We can write $T$ as matrix, which is a upper triangular matrix with diagonal $\{1,0,-1,0,1,0,-1,\cdots\}$, so I guess the answer is $\lfloor n/2 \rfloor+1$,but I can't prove it. I calculated $n=2$, and the answer is right.
 A: Write $ \mathcal{F}f(x) = \int_{\mathbb{R}} e^{-2\pi i x y} f(y) \, \mathrm{d}y $ for the Fourier transform, and let $\mathcal{M}f(x) = e^{-\pi x^2}f(x)$. Also let
$$ E_n = \{ f \in P_n : \text{$f$ is even} \}. $$
Then it is easy to check that:


*

*If $f \in P_n$ solves $Tf = f$, then $f \in E_n$.

*For $f \in E_n$, we have $Tf = \mathcal{M}^{-1}\mathcal{F}\mathcal{M}f$.

*Equip $E_n$ with the inner product $\langle f, g \rangle := \int_{\mathbb{R}} \overline{\mathcal{M}f(x)}\mathcal{M}g(x) \, \mathrm{d}x $. Then by the Plancherel's Theorem,
$$ \langle Tf, Tg \rangle
= \int_{\mathbb{R}} \overline{\mathcal{F}\mathcal{M}f(x)}\mathcal{F}\mathcal{M}g(x) \, \mathrm{d}x
= \int_{\mathbb{R}} \overline{\mathcal{M}f(x)}\mathcal{M}g(x) \, \mathrm{d}x
= \langle f, g \rangle, $$
and so, $T$ restricted to $E_n$ is an isometry with respect to this inner product.
This proves that $T$ restricted to $E_n$ is diagonalizable. So the geometric multiplicity of $1$ coincide the algebraic multiplicity of $1$, which is $\lfloor n/2\rfloor + 1$ as computed by other users.
