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A self-adjoint diagonally dominant square matrix $M$ with nonnegative diagonal is positive semi-definite. Does there exist a similar concept for integration kernels that define compact operators over, say, $L^2(\mathbb{R}^n)$?

Let me be more specific:

Suppose $M$ is a self-adjoint square matrix. $M$ is diagonally dominant if and only if, for all $i$, $$ |M_{ii}| \geq \sum_{j\neq i} |M_{ij}|. $$ $M$ is invertible. Suppose the diagonal satisfies $M_{ii}\geq 0$. Then $M$ is automatiaclly positive semidefinite. Thus, a simple criterion on the matrix elements determine whether $M$ is positive semidefinite -- a question which in general would involve computation of the whole spectrum of $M$.

Let us turn to integration kernels. Suppose $u:\mathbb{R}^n\times\mathbb{R}^n\rightarrow\mathbb{C}$ is (for simplicity assumed to be) a Schwartz function, such that we may define a compact operator $U$ over $L^2(\mathbb{R}^n)$ using $u$ as integration kernel, viz, $$ [U\phi](x) := \int_{\mathbb{R}^n} u(x,y)\phi(y) d^n y. $$ Assume $u(x,y) = \overline{u(y,x)}$ such that $U$ is self-adjoint. $u(x,y)$ is analogous to the matrix elements $M_{ij}$ of $M$.

Does there exist some sort of criterion for $u(x,y)$ analogous to $M$ being diagonally dominant, that guarantees $U\geq 0$? I.e., can we say something about whether $U$ is postive semidefinite by studying the behavior of $u(x,y)$ near the diagonal $x=y$? In general, positive semidefiniteness of $U$ is much harder to ascertain, i.e., we would have to study the spectrum of $U$.

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There's a notion of positive semi-definiteness for kernels, which is discussed in detail here:… – Branimir Ćaćić Jan 3 '13 at 14:23
So, is the question about the positivity of Hilbert-Schmidt operators? Hit it with the spectral theorem to decouple your kernel in that case. – Jonas Teuwen Jan 3 '13 at 15:17
@JonasTeuwen: Ah, no ... I have updated the question, should be much clearer now. – Jas Ter Jan 4 '13 at 12:07
@SimenK. But, I do not see understand why my decomposition would not be the way to proceed. Self-adjointness amounts to real eigenvalues and so on. – Jonas Teuwen Jan 4 '13 at 12:10

$\newcommand{\R}{\mathbf R}$ $\newcommand{\geq}{\geqslant}$ Let us start with the remark that as $u$ is Schwartz, $u$ is in $L^2(\R^d \times \R^d)$. Hence, the operator $U$ given by $$ \begin{align*} U : L^2(\R^d) &\mapsto L^2(\R^d)\\ v &\mapsto \int_{\R^d} u(x, y) v(y) \, \mathrm{d}y. \end{align*} $$ is bounded, and self-adjoint by the condition $u(x, y) = \overline{u(y, x)}$. $U$ is positive if $u \geq 0$ implies that $Uu \geq 0$.

We are now in a very good situation and indeed very similar to the one you state. We have a compact self-adjoint operator on a Hilbert space. Which means that there is a countable orthonormal basis of $L^2$ consisting of eigenvectors $u_n$ such that their eigenvalues $\lambda_n$ converge to $0$.

Nice, under good conditions (Mecer's theorem for instance) we can now see that $$u(x,y) = \sum_{n = 0}^\infty \lambda_n u_n(x) u_n(y).$$

And so, $$Uu(x) = \sum_{n = 0}^\infty \lambda_n \langle u_n, v\rangle u_n(x).$$

Is this the type of decomposition you are looking for?

Some additions: Let us see. As $u$ is Schwartz, we have that $U$ is a trace-class operator. That means that for our abstract expansion, by ordering such that the eigenvalues have the same sign from some $N$ on (by the spectral theorem) $$\sum_{n = 0}^\infty \lambda_n < \infty.$$ Additionally, the trace of $U$, $$\text{tr}\, U = \int_{\R^d} u(x, x) \, \text{d}{x} < \infty.$$

If $U$ is a positive operator, that is if $u \geq 0$ implies that $Uu \geq 0$, then then the self-adjointness states that $(\lambda)_n$ is not only real -due to the self-adjoint property- but additionally positive.

Question to you: What is the type of control you want to have? Pointwise on-diagonal control of the kernel? Traces?

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Unfortunately, no, I realize my question was not clear, so I edited the formulation. – Jas Ter Jan 4 '13 at 12:08
I do not understand how an abstract decomposition approach can help me ... Also, there may be some confusion: positive semidefiniteness of $U$ does not mean that $u\geq 0$. Is your statement "$U$ is positive if $u\geq 0$ implies that $Uu\geq 0$" equivalent to the usual definition of $U$ being positive semidefinite? – Jas Ter Jan 4 '13 at 13:19
On every inner product space every self-adjoint operator $P$is positive if for all non-zero $u$ we have $\langle Pu, u \rangle > 0$. – Jonas Teuwen Jan 4 '13 at 16:25
I agree with your post. I am not really aiming at solving a particular problem, more like exploring the theory. But I imagined some result like: suppose $u(x,y)$ near the diagonal behaves like blabla compared to the off-diagonal. Then, $U$ is positive semidefinite. I imagine the condition would compare the integral of $u(x,y)$ wrt $y$ for $|x-y|<\epsilon$ versus the same for for $\geq \epsilon$. – Jas Ter Jan 7 '13 at 12:29

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