Show operator is compact/symmetric

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In summary, the conversation discusses showing that the operator ##K## is compact by using a bounded sequence in ##L^{2}[0,1]##. The conversation also delves into finding eigenvalues and eigenfunctions of ##K##, with a focus on the constraints of the kernel. Ultimately, it is determined that there are only finitely many eigenfunctions with a nonzero eigenvalue, but there can be infinitely many with a zero eigenvalue. The relevance of the constraint ##\int_0^1 yf(y)\,dy = 0## for determining eigenvalues is left to be explored.
  • #1
member 428835
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.
 
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  • #2
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.
 
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  • #3
mathwonk said:
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
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
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
pasmith said:
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.
 

What does it mean for a show operator to be compact?

A show operator is said to be compact if the image of any bounded set is relatively compact. In simpler terms, this means that the operator maps bounded sets to sets that are relatively compact, or have a finite limit point.

What is the significance of a compact show operator?

A compact show operator has many important properties, such as being continuous, having a bounded inverse, and being a Fredholm operator. These properties make it a valuable tool in functional analysis and other areas of mathematics.

Can a non-compact show operator be symmetric?

Yes, a show operator can be both non-compact and symmetric. Compactness and symmetry are independent properties of a show operator, meaning that one does not imply the other. A symmetric operator is one that satisfies the condition (Tx,y) = (x,Ty) for all x and y in the domain of the operator.

How can one determine if a show operator is compact?

There are several criteria for determining if a show operator is compact, such as the Arzela-Ascoli theorem, the Riesz-Fischer theorem, and the Banach-Alaoglu theorem. These theorems provide necessary and sufficient conditions for a show operator to be compact.

What are some applications of compact show operators?

Compact show operators have many applications in mathematics, physics, and engineering. They are commonly used in solving integral equations, studying differential equations, and in functional analysis. They also have applications in signal processing, image processing, and data compression.

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