$V$ is the space of infinitely differentiable functions $f:\mathbb R \to \mathbb C$ which are periodic of $2\pi$, with the inner product $<f|g> = \int_0^{2\pi} \overline f(t)g(t)dt$. Let $L:V\to V$ be defined as $L=-d^2/dt^2$. It is known that $L$ is self-adjoint, $L=L^*$. How to show that the eigenvalues of $L$ are non-negative? And how to write $L=A^*A$ for an operator $A:V\to V$?
Show that the eigenvalues of this Hermitian operator are non-negative
functional-analysislinear algebra
Related Solutions
Let $\lambda \in \mathbb{R}$ be an eigenvalue of $T^* T$ and $v \neq 0$ an associated eigenvector. Then
$\lambda \langle v, v\rangle = \langle T^* T v, v\rangle = \langle Tv, Tv\rangle \geq 0$
implies $\lambda \geq 0$.
Note that the diagonal $\Delta=\{(x,x)\mid x\in[0,1]\}$ is a measure zero set, so it is not a priori clear what the meaning should be of a symbol such as $k(x,x)$. But we do have the following
Theorem: $K$ has an eigenbasis $(e_i)_{i\in\mathbb{N}}$ with corresponding eigenvalues $(\lambda_i)_{i\in\mathbb{N}}$. The function $q: [0,1]^2 \rightarrow \mathbb{C}$ defined by $q(x,y):=\sum_{i=1}^\infty \lambda_i e_i(x) \bar{e}_i(y)$ satisfies $k(x,y)=q(x,y)$ almost everywhere, and if we put $k(x):=q(x,x)=\sum_{i=1}^\infty \lambda_i \lvert e_i(x) \rvert^2 $, then $\mathrm{Tr }\, K = \int k(x)\,dx$.
Proof: Since $K$ is self-adjoint and compact, the spectral theorem ensures the existence of an ONB $(e_i)_{i\in\mathbb{N}}$ such that, for any $\psi\in L^2([0,1],\mathbb{C})$, we have $$ K\psi=\sum_{i=1}^\infty \lambda_i \langle e_i,\psi\rangle e_i. $$ Furthermore, since $K$ is Hilbert-Schmidt, we have $\sum_{i=1}^\infty \lvert \lambda_i\rvert^2<\infty$. Fixing representatives $e_i:[0,1]\rightarrow \mathbb{C}$ for the basis $(e_i)_{i\in\mathbb{N}}$, it follows that the series $$ q(x,y):=\sum_{i=1}^\infty \lambda_i e_i(x) \bar{e}_i(y) $$ is convergent in $L^2([0,1]^2,\mathbb{C})$, and since we have \begin{align} \int \bar{\phi}(x) k(x,y) \psi(y) \, dxdy & = \langle \phi , K\psi\rangle= \sum_{i=1}^\infty \lambda_i \langle \phi, e_i\rangle \langle e_i,\psi\rangle \\ &= \sum_{i=1}^\infty \lambda_i \int \bar{\phi}(x) e_i(x) \bar{e}_i(y) \psi(y)\, dxdy = \int \bar{\phi}(x) q(x,y) \psi(y)\, dxdy \end{align} for all $\phi,\psi\in L^2([0,1],\mathbb{C})$, it follows that $q=k$ almost everywhere.
Now we make use of the assumption that $K$ is trace class: We have $\sum_{i=1}^\infty \lvert \lambda_i\rvert<\infty$. Thus, we can define $k(x):=q(x,x)=\sum_{i=1}^\infty \lambda_i \lvert e_i(x)\rvert^2$, and then we have $k\in L^1([0,1],\mathbb{C})$. Furthermore, we have $$ \int k(x)\,dx = \sum_{i=1}^\infty \lambda_i \int \lvert e_i(x)\rvert^2\, dx = \sum_{i=1}^\infty \lambda_i = \mathrm{Tr }\, K, $$ finishing the proof.
This solution does not make use of non-negativity of K. To make the connection to the answer by Christian Remling, note that $K$ has a non-negative square root $A=\sqrt{K}$, and we have $$ A\psi=\sum_{i=1}^\infty \sqrt{\lambda_i} \langle e_i,\psi\rangle e_i. $$ Since $K$ is trace class, it follows that $A$ is Hilbert-Schmidt, and as above we find a kernel for $A$, namely $$ a(x,y)=\sum_{i=1}^\infty \sqrt{\lambda_i}e_i(x)\bar{e}_i(y). $$ Then we have $$ q(x,y) = \sum_{i=1}^\infty \lambda_i e_i(x)\bar{e}_i(y) = \int a(x,t)a(t,y) \, dt , $$ and in particular $k(x)=\int a(x,t)a(t,x) \, dt$. Here, it is also possible to remove the assumption that $K$ is non-negative, by noting that the polar decomposition $K=U\lvert K \rvert=(U\lvert K \rvert^{1/2})\lvert K \rvert^{1/2}$ affords us with a factorization of $K$ into a product of Hilbert-Schmidt operators.
Questions of this sort have received some attention in the literature, see for instance the paper 'Traceable Integral Kernels on Countably Generated Measure Spaces' by Brislawn. There is also a related question on MO, When is an integral transform trace class.
Brislawn, Chris, Traceable integral kernels on countably generated measure spaces, Pac. J. Math. 150, No.2, 229-240 (1991). ZBL0724.47014.
Best Answer
The associated form for $L$ is \begin{align} \langle Lf,f\rangle &= \int_{0}^{2\pi} -f''(t)\overline{f(t)}dt\\ &= -f'\overline{f}|_{0}^{2\pi}+\int_{0}^{2\pi}f'(t)\overline{f'(t)}dt \\ &= \int_{0}^{2\pi}|f'(t)|^2dt. \end{align} So, if $Lf=\lambda f$, then $\lambda\|f\|^2=\|f'\|^2$ forces $\lambda$ to be real and non-negative. Furthermore, $\lambda =0$ iff $f'=0$ or $f$ is constant.