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The integral operator $K:L^2 ([0,1]) \to L^2([0,1])$ is defined by $$(Ku)(x)=\int^1 _0 k(x,y)u(y) dy$$ Where $k(x,y)=min\{x,y\}$ for $0 \leq x, y \leq 1$

When finding the spectrum ok $K$ we let $Ku=\lambda u$

Then we get $$\lambda u(x)=\int^x _0 yu(y) dy + x \int^1 _x u(y) dy$$

How do we get this?

Furthermore if we want to show that $K$ is compact and self adjoint, we have to show that $$k(x,y)=\overline{k(y,x)}$$

In my solution it says $k(x,y)=\overline{k(y,s)}$ I am guessing the $s$ should be an $x$?

What does line represent? Is it closure?

How do we check if $k(x,y)=\overline{k(y,x)}$ holds true?

Lastly, how do we check if $k(x,y)$ is continuous?

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Al jabra
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  • The equation $\min {x,y } = \frac{x + y - |x - y|}{2}$ might be of help. Apply the equation then split the integral into the intervals $[0,x]$ and $[x,1]$. – Kayle of the Creeks May 01 '16 at 20:42
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    Break the interval of integration into $[0,x]\cup[x,1]$ and then look at the cases where $0 \le y \le x$ and where $x \le y \le 1$. Then you can apply the definition of $k$. – Disintegrating By Parts May 01 '16 at 20:43
  • @kayleofthecreeks where did you get that equation from? – Al jabra May 01 '16 at 20:57
  • By inspection. If $x \geq y$ then $ x + y - |x - y| = x + y - (x - y) = 2y$ and if $y > x$ then $x + y - |x - y| = x + y - (y - x) = 2x$. – Kayle of the Creeks May 02 '16 at 07:36
  • @Kayle of the Creeks : Question to a farmer: how do you count how many hens you have? Answer: Very simple : I count the number of legs and I divide by 2.. See the connection ? :) – Jean Marie May 02 '16 at 13:00

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Let us see in detail the operations; as $$\int^1 _0=\int^x _0+\int^1 _x $$

$$Ku=\lambda u(x)=\int^{y=x} _{y=0} min(x,y) u(y) dy + \int^{y=1} _{y=x} min(x,y) u(y) dy$$

But, in the first integral, where $0\leq y \leq x$, the smallest among the two numbers $x$ and $y$ is $y$ ; this is why, in this case, $min(x,y)$ is replaced by $y$. A symmetrical situation happens in the second integral:

$$\lambda u(x)=\int^x _0 yu(y) dy + \int^1 _x x u(y) dy$$

$$\lambda u(x)=\int^x _0 yu(y) dy - x \int^x _1 u(y) dy \ \ \ (1)$$

The next step would be to derive (1) twice in order to get a second order differential equation for $u$, and then solve it. See the question I have been asking : Looking for examples of Discrete / Continuous complementary approaches

Jean Marie
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