Show operator is compact/symmetric

  • #1
joshmccraney
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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.
 
Last edited:

Answers and Replies

  • #2
mathwonk
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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.
 
Last edited:
  • #3
joshmccraney
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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?
 
  • #4
mathwonk
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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.
 
  • #5
pasmith
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If [itex]\int_0^1 f(y)\,dy = 0[/itex] then [itex]K[f][/itex] is a constant. Since [tex]
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
[/tex] we can construct for each such [itex]f[/itex] a function [tex]
g : [0,1] \to \mathbb{R} : x \mapsto f(x) - (12x - 6)K[f]
[/tex] which is then in the kernel of [itex]K[/itex].

I leave you to consider the case [itex]\int_0^1 yf(y)\,dy = 0[/itex], and whether every kernel function must satisfy at least one of those constraints.
 
  • #6
joshmccraney
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98
If [itex]\int_0^1 f(y)\,dy = 0[/itex] then [itex]K[f][/itex] is a constant. Since [tex]
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
[/tex] we can construct for each such [itex]f[/itex] a function [tex]
g : [0,1] \to \mathbb{R} : x \mapsto f(x) - (12x - 6)K[f]
[/tex] which is then in the kernel of [itex]K[/itex].

I leave you to consider the case [itex]\int_0^1 yf(y)\,dy = 0[/itex], 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.
 

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