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Consider the (compact) operator $T:C([0,1],\mathbb{R})\rightarrow C([0,1],\mathbb{R})$ s.t. \begin{equation} T(f)(x)=\int_0^1\min\{x,y\} f(y)dy \; . \end{equation} How could one find its spectrum? I have tried to impose $Tf-\lambda f \equiv 0$ but then I couldn't solve anything...

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    $\begingroup$ If you look for the keywords such as: operator min kernel covariance brownian, you will find a number of answers. E.g. here. $\endgroup$ – user66081 Aug 20 '16 at 5:56
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A fundamental fact about this operator is that its inverse is the opposite of the second derivative. I show it in (Looking for examples of Discrete / Continuous complementary approaches).

What I give here is not an answer for the continuous operator "min" but for its discrete equivalent as a $n \times n$ matrix $M_n$ (or $M$) defined by

$$M_{i,j}=min(i,j) \ \ \ \ 1 \leq i,j \leq n.$$

Two reasons for studying this issue:

  • it helps to understand the behaviour of the continuous operator "min" .

  • it is essential in the implementation of this continuous operator on a digital computer.

In the same reference, I show also that the inverse of the discrete operator "min" as a $n \times n$ matrix still is the opposite $D_n$ of the discrete analog of the second derivative.

In my answer to a question (https://math.stackexchange.com/q/1880704), I show that the spectrum of $D_n$ is the set of

$$\lambda_p=4 \left(\sin\dfrac{p \pi}{n+1}\right)^2 \ \ (p=1 \cdots n)$$

Thus the spectrum of $M$ is the set of all $\dfrac{1}{\lambda_p}$.

Remark : $M_n$ and $D_n$, having positive eigenvalues, are symmetric positive definite matrices.

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  • $\begingroup$ @William Tomblin You haven't made any comment on my answer... which was "alone in its category" :) $\endgroup$ – Jean Marie Jun 8 '17 at 0:46
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It is fairly easy to solve the equation $Tf=\lambda f$ for $\lambda \neq 0$.. We have $\int_o ^{x} yf(y)dy +x\int_x^{1} f(y)dy=\lambda f(x)$. Observe that continuity of $f$ implies differentiability of the left side, hence of $f$ itself. In fact, by induction, we see that $f$ has derivatives of all orders. Differentiate once to get $xf(x)+\int_x^{1} f(y)dy-xf(x)=\lambda f'(x)$. Differentiate again to get $\lambda f''(x)=-f(x)$. Note that $\lambda$ is real because $f$ is real valued. The general solution of the equation $\lambda f''(x)+f(x)=0$ can be written as linear combination of sine and cosine functions if $\lambda >0$ and linear combination of $e^{cx}$ and $e^{cx}$, where $c=\frac 1 {\sqrt {\lambda}}$, if $\lambda <0$. Now remember that (from the steps leading to the differential equation) we have $f(0)=0$ and $f'(1)=0$. Now you can write exactly what the eigen values are and even write down the eigen functions.

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