I was trying to do this exercise and I'm wondering if I figured it out well:

I have $\mathcal{H} := L^2(0,1)$ and $T$ the operator with integral kernel $K(x,y) = \min\{x,y\}$, $x,y \in [0,1]$. I have to show that $T$ is compact and self-adjoint.

To show that is compact I was thinking to say that because $\min\{x,y\} \in [0,1]$ then

\begin{equation} \dim(\operatorname{Im}T) = 1 \end{equation}

(The self adjointness I think is trivial..)So T belongs to finite rank operators and so it is compact. (Is this correct?) Then it asks me to find eigenvalues and eigenvectors of $T$ and here I really don't know how to proceed...

  • $\begingroup$ When you claim the dimension is 1, you mean to claim that there is a function $f$ such that for any function $g\in L^2$, there is a constant $\lambda=\lambda(g)$ such that $\int_0^1 K(x,y)g(y)dy = \lambda f(x)$? $\endgroup$ – Calvin Khor Jan 5 at 19:17
  • $\begingroup$ Actually I meant that $ImT = < 1 >$. But I don't know if it's right.. $\endgroup$ – James Arten Jan 5 at 19:20
  • $\begingroup$ So I think in what I wrote, you'd be claiming that the function $f$ is identically 1. No, this is not true, if you take $g=1\in L^2$ then $\int_0^1 \min(x,y) \cdot 1 dy $ is not constant in $x$. $\endgroup$ – Calvin Khor Jan 5 at 19:23
  • $\begingroup$ Yea sorry I didn't mean < 1 > because if I consider $\int_0^{1} min\{x,y\}f(y)\,dy$ it will be a function of $x$. So the $ImT$ will be generated by all possible linear combinations of $x$, is this right? $\endgroup$ – James Arten Jan 5 at 19:27
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    $\begingroup$ Please observe in desmos.com/calculator that $T1$ is not linear $\endgroup$ – Calvin Khor Jan 5 at 19:27

We have $$ \int_{(0,1)^2} |k(x,y)|^2\ \mathsf d(x\times y) = \int_0^1\int_0^1 (x\wedge y)^2\ \mathsf dx\ \mathsf dy \leqslant \int_0^1\int_0^1\ \mathsf dx\ \mathsf dy = 1 <\infty, $$ so T is a Hilbert-Schmidt operator and hence is compact.

  • $\begingroup$ Thanks! What about eigenvalues/eigenvectors? $\endgroup$ – James Arten Jan 5 at 20:42

$T$ is a Hibert-Schmidt operator because $\min\{x,y\} \in L^2([0,1]\times[0,1])$. $Tf$ may be written as \begin{align} Tf & = \int_{0}^{1}K(x,y)f(y)dy \\ & = \int_{0}^{1}\min\{x,y\}f(y)dy \\ & = \int_0^xyf(y)dy+x\int_x^1 f(y)dy \end{align} If $Tf=\lambda f$ for some $f\in L^2$ and $\lambda\in\mathbb{C}\setminus\{0\}$, then the above implies that $Tf$ is equal a.e. to a continuous function on $[0,1]$. Hence, $f$ is equal a.e. to a continuous function. So $Tf$ is continuously differentiable. So, assume without loss of generality, that $f$ is continuous. Then $(Tf)(0)=0$ and $(Tf)'$ exists with $$ \lambda f'= (Tf)'=xf(x)-xf(x)+\int_x^1f(y)dty=\int_x^1 f(y)dt $$ So $f$ is $C^2$, $f(0)=0$, $f'(1)=0$ and $\lambda f''=-f$ for every eigenfunction with eigenvalue $\lambda\ne 0$. The eigenfunctions with non-zero eigenvalues are, therefore, constant multiplies of $$ f_n = \sin(n\pi x/2),\;\;\; n=1,3,5,7,\cdots, \\ \lambda_n = \frac{2}{n\pi}. $$

The adjoint of $\int_0^x$ is $\int_x^1$. And the adjoint of $M_x$ (multiplication by $x$) is $M_x$. So $T$ is selfadjoint because \begin{align} T &= \left(\int_0^x\right)M_x+M_x\left(\int_x^1\right)\\ &=\left(\int_0^x\right)M_x+M_x^*\left(\int_0^x\right)^* = T^*.\end{align}


Since we have the integral identity $$ g:= Tf = \int_0^x yf(y) dy + x\int_x^1 f(y) dy $$ Then $g\in H^1$, and since $g' = \int_x^1 f(y) dy$, actually $g \in H^2$, with $$ -g'' = f$$ So we actually have a Poisson equation, but with boundary conditions $g(0)=0,g(1) = \int_0^1 yf(y) dy$. By setting $\tilde g = g - x \int_0^1 yf(y) dy$, we notice that we are equivalently looking for the weak solutions in $H^1_0$ to the Poisson equation with Dirichlet boundary conditions $$ -\tilde g'' = f \text{ on } (0,1),\quad \tilde g(0)=\tilde g(1)=0,\quad f\in L^2 $$ So if you already knew that the solution operator $f\mapsto \tilde g$ for this 1D Poisson equation was compact($H^1_0 \subset\subset L^2)$ and self-adjoint, we're done (the difference $ g-\tilde g= x\int_0^1 y f(y) dy$ is finite rank and self-adjoint).


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