 #1
joshmccraney
Gold Member
 2,253
 143
 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 ##xx'+yy' < \delta##. Therefore, ##\{ Kf_{n}\}## is a sequence of continuous functions for which
$$
Kf_{n}(x)Kf_{n}(x') \le \int_{0}^{1}k(x,y)k(x',y)f_{n}(y)\,dy \\
\le \epsilon \int_{0}^{1}f_{n}(y)\,dy \le \epsilon\1\\f_{n}\
\le M\epsilon,\;\;\; xx' < \delta
$$
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.
$$
Kf_{n}(x)Kf_{n}(x') \le \int_{0}^{1}k(x,y)k(x',y)f_{n}(y)\,dy \\
\le \epsilon \int_{0}^{1}f_{n}(y)\,dy \le \epsilon\1\\f_{n}\
\le M\epsilon,\;\;\; xx' < \delta
$$
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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