- #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.

$$

|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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