Show operator is compact/symmetric

[joshmccraney]

Homework Statement:

Consider the operator $K:L_2[0,1]\to L_2[0,1]$ given by $Kf(x) = \int_0^1(x+y)f(y)\, dy$
1) Prove that $K$ is a compact symmetric operator.
2) Find all eigenvalues of $K$.
3) Manually verify that eigenvectors of $K$ corresponding to different eigenvalues are pairwise orthogonal.

Relevant Equations:

Nothing comes to mind

1) To show that $K$ is compact let $\{ f_{n} \}_{n=1}^{\infty}$ be a bounded sequence in $L^{2}[0,1]$ with $\|f_{n}\| \le M$. For every $\epsilon > 0$, there exists $\delta > 0$ such that $|k(x,y)-k(x',y')| < \epsilon$ whenever $|x-x'|+|y-y'| < \delta$. Therefore, $\{ Kf_{n}\}$ is a sequence of continuous functions for which
$$
|Kf_{n}(x)-Kf_{n}(x')| \le \int_{0}^{1}|k(x,y)-k(x',y)||f_{n}(y)|\,dy \\
\le \epsilon \int_{0}^{1}|f_{n}(y)|\,dy \le \epsilon\|1\|\|f_{n}\|
\le M\epsilon,\;\;\; |x-x'| < \delta
$$
where $k(x,y) = x+y$. Thus $\{ Kf_{n} \}$ is an equicontinuous family of continuous functions on $[0,1]$. So, there exists a subsequence $\{ Kf_{n_{k}}\}$ that converges uniformly to a continuous function $g$. Since uniform convergence implies convergence in $L^{2}[0,1]$, it follows that $\{ Kf_{n_{k}}\}$ converges in $L^{2}[0,1]$. Therefore $K$ is compact because the image of a bounded sequence always contains a convergent subsequence.

2) no clue. I'm thinking $\det (\lambda I - K) = 0$? Then $\int_0^1(x+y) f(y)\, dy = \lambda f(x)$? It seems obvious to me that $f$ must be a linear function, so that $f(x) = a x + b$. Then I think $$\int_0^1(x+y)(ay+b)\,dy = \lambda(ax+b)\implies\\
a/3 + b/2 + (a/2 + b) x = \lambda b + \lambda a x.$$ Weighting equations implies $\lambda_1 = 1/6 (3 - 2 \sqrt 3)$ when $b = -(a/\sqrt 3 )$ or $\lambda_2 = 1/6 (3 + 2 \sqrt 3)$ when $b = (a/\sqrt 3 )$. Then our eigenvalues are determined and eigenfunctions are determined up to the constant $a$. Is the right so far? If so, what's next?

3) I need 2) to even attempt.

 

[mathwonk]

I suggest simplifying your formula for (Kf)(x). It becomes obvious that (Kf)(x) is always linear, with linear coefficient the integral of f and constant term the integral of y. f(y). Hence you are right that any eigenfunction f must also be linear. Then just compute as you have done to find all eigenvalues (I have not checked fully your computations). (The determinant seems useless in infinite dimensions.)

I do not know what you mean by "weighting equations", but setting coefficients equal on both sides seems to give that f = ax+b is an eigenfunction iff 3b^2 = a^2, and the corresponding eigenvalue is (a+2b)/2a = (2a+3b)/6b. Is that what you got? oh yes, indeed then b = a/sqrt(3), so probably I agree.

You seem to have solved 2).

wait, what am i doing wrong? shouldn't a compact symmetric operator have a lot of eigenfunctions? maybe i am being careless about assuming ab ≠ 0. its dinnertime.

or maybe i need to determine the kernel of the operator. that looks more interesting. i seem to have been assuming the eigenvalue was ≠0.

 

[joshmccraney]

wait, what am i doing wrong? shouldn't a compoact symmetric operator hve a lot of eigenfunctions? maybe i am being careless about assuming ab ≠ 0. its dinnertime.

I think only two is fine. I just googled and found an example here on page 3:

http://www.maths.qmul.ac.uk/~ig/MAS214/int-ops-lect.pdf
I didn't really read their technique but it looks like they recover only two eigenvalues. What do you think?

 

[mathwonk]

finitely many eigenvalues is cool, but doesn't a compact symmetric operator on L2 have an infinite number of eigenfunctions?

Oh yes, the kernel seems to be the key. Note that for any smooth function F with F(0) = F(1) = integral of F over [0,1], the derivative F' seems to be in the kernel. These are very easy to construct with F(0) = 0. e.g. F(x) = sin(2nπx). but i have not checked this.

Indeed the general theory seems to say that there are only finitely many eigenfunctions sharing the same ≠0 eigenvalue, but there can be infinitely many with eigenvalue zero. But this is not my game. It seems kind of fun though, and I used to like it back in the day.

 

[pasmith]

If $\int_0^1 f(y)\,dy = 0$ then $K[f]$ is a constant. Since $$K\left[ 12x - 6 \right] = x \int_0^1 \left(12y - 6\right)\,dy + \int_0^1 12y^2 - 6y\,dy = x \left( 6 - 6 \right) + \left(4 - 3\right) = 1$$ we can construct for each such $f$ a function $$g : [0,1] \to \mathbb{R} : x \mapsto f(x) - (12x - 6)K[f]$$ which is then in the kernel of $K$.

I leave you to consider the case $\int_0^1 yf(y)\,dy = 0$, and whether every kernel function must satisfy at least one of those constraints.

 

[joshmccraney]

If $\int_0^1 f(y)\,dy = 0$ then $K[f]$ is a constant. Since $$K\left[ 12x - 6 \right] = x \int_0^1 \left(12y - 6\right)\,dy + \int_0^1 12y^2 - 6y\,dy = x \left( 6 - 6 \right) + \left(4 - 3\right) = 1$$ we can construct for each such $f$ a function $$g : [0,1] \to \mathbb{R} : x \mapsto f(x) - (12x - 6)K[f]$$ which is then in the kernel of $K$.

I leave you to consider the case $\int_0^1 yf(y)\,dy = 0$, and whether every kernel function must satisfy at least one of those constraints.

Can you expound a little please? I'm confused what relevance $\int_0^1 yf(y)\,dy = 0$ has for determining eigenvalues.