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The space $C$ of continuous functions $f(u)$ on the interval $[0, 1]$ is one of many infinite-dimensional analogues of $\mathbb{R}^n$ , and continuous functions $A(u, v)$ on the square $0 \leq u, v \leq1$ are infinite-dimensional analogues of matrices. The integral $$A \cdot f=\int_0^1 A(u,v)f(v) dv$$ is analogous to multiplication of a matrix by a vector.

This problem treats the integral as a linear operator. For the function $A = u + v$, determine the image of the operator explicitly. Determine its nonzero eigenvalues, and describe its kernel in terms of the vanishing of some integrals.

So I replace $A(u,v)$ by $u+v$. Be obtain:

$u\int_0^1 f(v)dv + \int_0^1 vf(v) dv$ but I don't really see how to continue.

Can someone give me hints for the various questions ?

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  • $\begingroup$ What is the subspace of functions that you can write in the form $g = A\cdot f$? That subspace is the image of the operator. If $\lambda$ is a nonzero eigenvalue, there is an $f \neq 0$ with $A\cdot f = \lambda\cdot f$. In particular, $\lambda f$, and hence $f$ itself, must belong to the image. When is $A\cdot f = 0$? $\endgroup$ – Daniel Fischer Oct 13 '13 at 21:45
  • $\begingroup$ @DanielFischer Thank you. So I will have $g(u)=u\int_0^1 f(v)dv + \int_0^1 vf(v) dv$ but how do I continue from there ? $\endgroup$ – user43418 Oct 13 '13 at 21:48
  • $\begingroup$ No, not $g(v)$. It's $g(u)$. $\endgroup$ – Daniel Fischer Oct 13 '13 at 21:50
  • $\begingroup$ @DanielFischer Yes sorry (edited). But I don't understand what can we do more.. $\endgroup$ – user43418 Oct 13 '13 at 21:51
  • $\begingroup$ Look a bit at it. You should be able to see what form any function in the image has. Once you saw that, the further things aren't very hard. $\endgroup$ – Daniel Fischer Oct 13 '13 at 21:55
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For a continuous function $A \colon [0,1]\times [0,1] \to \mathbb{R}$, we consider the integral operator

$$T_A \colon f \mapsto \left( u \mapsto\int_0^1 A(u,v)f(v)\,dv\right).$$

We will look at the case $A(u,v) = u+v$ closer. We see that the general result of applying the operator $T_A$ in that case is

$$T_A(f)(u) = \int_0^1(u+v)f(v)\,dv = u\cdot \int_0^1 f(v)\,dv + \int_0^1 vf(v)\,dv.\tag{1}$$

So every function we can obtain by applying $T_A$ has the form $g(u) = a\cdot u + b$ for some constants $a,b \in \mathbb{R}$. We need to check whether every such function is in the image of $T_A$, or the image is a proper subspace of the space of linear polynomials. Choosing $f_0(v) = 2v - 1$, we have $\int_0^1 f_0(v)\,dv = 0$, so $T_A(f_0)(u) = \int_0^1 v(2v-1)\,dv = \frac23-\frac12 = \frac16$, and we see that every constant function can be obtained. Choosing $f_1(v) = 1$, we obtain $T_A(f_1)(u) = u + \frac12$. So we have

$$\operatorname{im} T_A = \{ au + b : a, b \in \mathbb{R}\}.$$

From the right hand side of $(1)$, we can directly read off that

$$\ker T_A = \left\lbrace f \in C : \int_0^1 f(v)\,dv = 0, \int_0^1 vf(v)\,dv = 0\right\rbrace.$$

An eigenfunction to a nonzero eigenvalue must necessarily be contained in the image of $T_A$, so we need to see for which $f(v) = av + b$ we have $T_A(f)(u) = \lambda f(u)$ for some $\lambda \neq 0$. The image of $T_A$ is a finite-dimensional space (two dimensional, concretely), so we can describe the operation of $T_A$ on its image by a matrix. Plugging the general form of $f$ into $(1)$, we find

$$T_A(av+b)(u) = u\cdot \left(\frac{a}{2} + b\right) + \left(\frac{a}{3} + \frac{b}{2}\right),$$

so it remains to find the eigenvaules and eigenvectors of the matrix

$$M_A = \begin{pmatrix} \frac12 & 1\\ \frac13 & \frac12\end{pmatrix}.$$

Its characteristic polynomial is $\chi_{M_A}(X) = X^2 - X - \frac{1}{12} = \left(X - \frac12 - \frac{1}{\sqrt{3}}\right)\left(X - \frac12 + \frac{1}{\sqrt{3}} \right)$.

The eigenvalues are $\frac12 + \frac{1}{\sqrt{3}}$ and $\frac12 - \frac{1}{\sqrt{3}}$, and corresponding eigenvectors are

$$\begin{pmatrix}\sqrt{3}\\ 1 \end{pmatrix},\qquad \begin{pmatrix}\sqrt{3}\\ -1 \end{pmatrix}.$$

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  • $\begingroup$ Shouldn't it be $M_A= \begin {pmatrix} \frac{1}{2} & \frac{1}{3}\\ 1 & \frac{1}{2} \end{pmatrix}$. I know it's the same thing since $\det(M-\lambda I)=\det(M^t-\lambda I)$ (but convention-wise...) $\endgroup$ – user43418 Oct 14 '13 at 19:20
  • $\begingroup$ That depends on which convention one has learned, and how one has ordered the basis. The one I learned and use is that the $j$-th column contains the coordinates of the image of the $j$-th basis vector. With the basis $(v,1)$, the coordinates of $T_A(v) = \frac12 u + \frac13$ are $\frac12$ and $\frac13$. If you ordered the basis the other way round, or used a different convention, you (may) need to transpose [for the same convention with the different ordering, you also need to swap the diagonal elements, but those are the same here, so it comes out to be a transpose]. $\endgroup$ – Daniel Fischer Oct 15 '13 at 9:28

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