# Integral Operator in $L^2$

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

$$$$\dim(\operatorname{Im}T) = 1$$$$

(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...

• 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 $\int_0^1 K(x,y)g(y)dy = \lambda f(x)$? – Calvin Khor Jan 5 at 19:17
• Actually I meant that $ImT = < 1 >$. But I don't know if it's right.. – James Arten Jan 5 at 19:20
• 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 $x$. – Calvin Khor Jan 5 at 19:23
• 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? – James Arten Jan 5 at 19:27
• Please observe in desmos.com/calculator that $T1$ is not linear – 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.
$$T$$ 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).