6
$\begingroup$

I need a small hint for solving this problem please.

Problem:
Consider the operator $ T: C([ 0,1 ])\to C([ 0,1 ]) $ which is defined as follows $$ T(u)(x)=\int_{0}^{x^{2}}u(\sqrt s)ds. $$ Prove that for each $\lambda\in \mathbb{R},$ there does not exist $ u\in C([ 0,1 ]) $ different from null function, such that $ Tu=\lambda u. $

What I have done: I tried to test the proposition with some examples. For instance, I put $ u(x)=x $ or $ u(x)=2x^{2}. $ And I found it without having any problems which contradicts with the above. Please let me know if there is something which is helpful.

$\endgroup$
  • $\begingroup$ Say u is x^2 then Tu is x^4/2, which is not a constant multiple of x^2. $\endgroup$ – Ian Jul 28 '17 at 21:35
5
$\begingroup$

Hint Proceed in two steps, the strategy being to find all the functions $u$ such that $T(u)=\lambda.u$ and then conclude.

First step (Find a condition - with derivatives - that solutions of $T(u)=\lambda u$ must satisfy and find one of them) If you set $$F(x):=\int_{0}^x u(\sqrt{s})ds\ .$$ Then $T(u)=F(x^2)$. If $u$ is s.t. $$T(u)=\lambda u\qquad (1),$$ taking derivatives, one gets. $$ \lambda u'(x)=T(u)'=2xF'(x^2)=2x.u(x)\qquad (2) $$ One first supposes $\lambda\not=0$. Then if $u$ satisfies (1), it satisfies (2) and $$ u'=\frac{2x}{\lambda}.u\qquad (3) $$ then, we have $u=e^{\frac{x^2}{\lambda}}$ as particular solution (call this solution $Z_\lambda$).

Second step (Find all solutions of $T(u)=\lambda u$ must satisfy and conclude) Consider any solution of $T(u)=\lambda u$. We form $y=(Z_\lambda)^{-1}u=e^{-\frac{x^2}{\lambda}}u$ and compute. As it must satisfy (3), one has $$ y'=(-\frac{2x}{\lambda})e^{-\frac{x^2}{\lambda}}u+e^{-\frac{x^2}{\lambda}}u'=(-\frac{2x}{\lambda})e^{-\frac{x^2}{\lambda}}u+e^{-\frac{x^2}{\lambda}}(\frac{2x}{\lambda})u=0 $$ then $y=c$ constant and $u=c.e^{\frac{x^2}{\lambda}}$. Now $$ T(u)=\int_{0}^{x^2} u(\sqrt{s})ds=c.\int_{0}^{x^2} e^{\frac{s}{\lambda}}ds=c.[\lambda.e^{\frac{s}{\lambda}}]_0^{x^2}=\lambda.c.(e^{\frac{x^2}{\lambda}}-1) $$ if $T(u)=\lambda.u$ this forces $\lambda.c=0$ and then $c=0$ hence the claim.

For $\lambda=0$, from (2), one gets $2x.u(x)=0$ and then $u\equiv 0$.

Remark (After Ranc's post) The operator $T$ is the conjugate of a Volterra operator (but not, in the usual sense, a Volterra operator). The conjugacy factor being the isometry $u\to U(u)$ with $U(u)(x)=u(x^2)$ ($T=UWU^{-1}$).

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ First off, thanks. $F'(x^{2})=2xu(x)$ then $u= e^{\frac{4x^{3}}{3\lambda}}. $ Any way, I could not understand the contradiction. $\endgroup$ – Hamit Jul 29 '17 at 7:09
  • 1
    $\begingroup$ $F'(X)=u(\sqrt{X})$ then, $F'(x^2)=u(x)$ and one has, in turn, $\lambda u'=2xu$, then the standard technique, for equations like $u'=m(x)u$ is to find a particular solution (exponential) $e^{M(x)}$ where $M'(x)=m(x)$ and to use it as counter-term. If something remains unclear, do not hesitate, all must/should become cristal clear. $\endgroup$ – Duchamp Gérard H. E. Jul 29 '17 at 7:20
  • $\begingroup$ @HamedBaghalGhaffari Thanks. Do not hesitate however if something remains unclear. $\endgroup$ – Duchamp Gérard H. E. Jul 30 '17 at 14:19
  • $\begingroup$ Thank you. It seems that I have to practice more in order to be accustomed to solve math problems. $\endgroup$ – Hamit Jul 31 '17 at 9:59
  • $\begingroup$ @HamedBaghalGhaffari You're right, the successful recipe being : practice + interaction. Good luck ! $\endgroup$ – Duchamp Gérard H. E. Jul 31 '17 at 13:31
2
$\begingroup$

I would like to suggest a solution that is more functional-analysis oriented, which is based on analysis of the spectrum of Volterra operators. This can be finished in a sentence:

It turns out $T$ is Volterra operator from $C([0,1])$ to itself, so the spectrum is $\{0\}$, but $0$ is not an eigenvalue.

Ofcourse, this is not a hint, nor it is easy, but I decided it is reasonable to show it, since you already got many good hints and answers. I will outline most of the proof, a major part (Neumann series THM) is omitted, but it is not difficult if you are familiar with some operator theory over Banach spaces.

Proof: Note $Tf(x) = \int_0 ^{x^2} f(\sqrt{y}) \mathrm{d}y = \int_0 ^x 2 f(\tau)\tau \mathrm{d}\tau$ which is to say that $T$ is a Volterra operator (and we can say alot about these). Recall that Neumann THM about operator series: Suppose $T\colon X \rightarrow X$ is an operator defined on a Banach space $X$, then if $\sum_{n=0}^\infty\| T^n \| < \infty$ we have $I-T$ is invertible, and its inverse is $\sum_{n=0}^\infty T ^n$.

Let us evaluate $\|T^n\|$, (recall $x\in(0,1)$): $$|Tf(x)|=\left| \int_0 ^x 2\tau f(\tau) \mathrm{d}\tau \right| \leq \int_0 ^x |2 f(\tau)| \mathrm{d}\tau \leq \int_0^x2\|f\|_\infty \mathrm{d}\tau=2\|f\|_\infty x $$ which gives a point-wise bound for $Tf(x)$ and also shows $\|T\| \leq 2$. Let us evaluate $T^2$ in a similar manner:

$$|T^2f(x)|= \left| \int_0^x 2\tau \cdot Tf(\tau) \mathrm{d}\tau \right| \leq \int_0^x |2\tau \cdot Tf(\tau) |\mathrm{d}\tau\leq 2^2\|f\|_\infty \int_0^x\tau \mathrm{d}\tau = 2^2 \|f\|_\infty \frac{x^2}{2}$$

which again gives a point-wise bound $|T^2f(x)| \leq 2\|f\|_\infty x^2$ and $\|T^2\| \leq 2\|f\|_\infty$. Inductive reasoning shows $$|T^nf(x)| \leq 2^n \|f\|_\infty \cdot \frac{x^{n}}{n!}$$ that gives $\|T^n\| \leq 2^n/(n-1)!$, so the series converge and $I-T$ is invertible, and $T-I$ is invertible.

Take $\lambda \in \mathbb{C} \setminus{0}$, and look at $T-\lambda I$. Is it invertible? Yes. $T-\lambda I = \lambda ( \lambda^{-1} T -I)$, so it is enough to show $ \lambda^{-1} T -I$ is invertible, but it is (since $\lambda^{-1}T$ is Volterra operator).

Since the spectrum $\sigma(T)$ is not empty - it must contain the point $\{0\}$, and we are left to determine if $0$ is an eigenvalue.

Although not immediate, showing $0$ is not an eigenvalue is easy, and I take the privilige to stop here.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ [which is to say that $T$ is a Volterra operator]---> it seems that it is not a Volterra operator in the usual sense (which is $u\to \int_{0}^xu(s)ds$) but the composite of it with $u\to 2x.u$ (which, in general affects the spectrum). How can you repair your proof ? $\endgroup$ – Duchamp Gérard H. E. Jul 30 '17 at 4:05
  • $\begingroup$ O.K, I think you can repair reusing your majoration of $T^n$ to majorate similarly $\lambda^{-n}T^n$, conclude (and withdraw the claim that $T$ is Volterra, unless I miss something :) $\endgroup$ – Duchamp Gérard H. E. Jul 30 '17 at 4:59
  • $\begingroup$ @DuchampGérardH.E. The definition I remember is $Tf (x)= \int_0^x k(x,y) f(y)\mathrm{d}y$ is Volterra, but I see that Wikipedia en.wikipedia.org/wiki/Volterra_operator offers another, more simple definition (similar to yours). However, correct me if I am wrong, other than calling it Volterra operator, everything in the proof works out fine? It is true that the spectrum of (what you and Wikipedia call) Volterra operators is just $\{0\}$, but I showed the same is true for the operator given in the question. $\endgroup$ – Ranc Jul 30 '17 at 6:32
  • $\begingroup$ @DuchampGérardH.E. as for the majoration, for any $\lambda \ne 0$ we simply have $\|\lambda^{-1} S\| = |\lambda^{-1}| \|S\|$, so the majoration goes right through with a finite multiplicative term. (I don't see why there would appear $\lambda^{-n}$). $\endgroup$ – Ranc Jul 30 '17 at 6:37
  • $\begingroup$ @DuchampGérardH.E. you made me check my books :): Basic Classes of Linear Operators by Gohberg, Goldberg, Kaashoek , and Functional Analysis: An Introduction by Eidelman, Milman, Tsolomitis; both define Volterra operators to be Tf(x)=\int_0^xk(x,y)f(y)\mathrm{d}y. I find the definition in Wikipedia to be too simple, for instance, it does not provide the useful generality for theory of integro-differential equations. $\endgroup$ – Ranc Jul 30 '17 at 6:47
2
$\begingroup$

It's much easier than solving ODE:

Note that $ T(u)(x)=2\int_{0}^{x} su (s) ds $,

Hence, for each $0<\lambda\leq 1$ you have $$ \sup_{x\in [0,\lambda] }| T (u)(x)|\leq 2\lambda sup_{x\in [0,\lambda]}|u (x)| $$ which shows that any eigenvalue must be smaller than $2\lambda $ for all $\lambda\in (0,1]$. In other words, zero is the only possible eigenvalue.

To see that zero isn't an eigenvalue, note that $[T(u)(x)]'=2xu (x) $. Hence, if $ T (u)\equiv 0$, then $ 2x u (x)=0$ for all $ x $, so $ u (x) =0$ for all $ x> 0$ and, by continuity, also for $ x=0$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I do not understand your inequality (*) (I added this in order to interact) don't you have a factor 2 ? $\endgroup$ – Duchamp Gérard H. E. Jul 30 '17 at 9:43
  • $\begingroup$ I had forgotten this factor in my first equation (you now fixed this), so yes, I also was missing this factor in inequality (*) $\endgroup$ – Bananach Jul 30 '17 at 12:50
1
$\begingroup$

Note:

$$\int\limits_0^{x^2} u(\sqrt{s})ds = \int\limits_0^{x^2}\sqrt{s}ds = \frac{2}{3}\Big[s\sqrt{s}\Big]^{x^2}_0 = \frac{2}{3}x^3$$

so taking $u(x) = x$ isn't a counterexample.

EDIT: In fact for all $u(x) = x^n$ we would have $$T(u)(x) = \int s^{n/2} ds \sim (x^2)^{n/2 + 1} \sim x^{n+2}$$

so the power will differ by $2$. I'd try constructing general Taylor series and showing that can't be a solution for any set of coefficients.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ $\frac{2}{3} x^{3}=\lambda x $ so if $\lambda =1, $ then $u(x)=x$ is an answer for the question which is a contradiction. $\endgroup$ – Hamit Jul 28 '17 at 18:49
  • 1
    $\begingroup$ @HamedBaghalGhaffari I'm sorry but what? If $\lambda = 1$ then you got $\frac{2}{3} x^3 = x$ which for sure isn't true for all $x \in [0,1]$. Just plug in $x=1$. $\endgroup$ – Piotr Benedysiuk Jul 28 '17 at 18:52
  • $\begingroup$ Good point. Thank you. I did not pay attention to the domain. Now, I have to find a proof for the problem... $\endgroup$ – Hamit Jul 28 '17 at 18:57
  • $\begingroup$ @DuchampGérardH.E. Could you post an answer to this question by solving this DE? I never solved one with integrals and nontrivial ones at that. $\endgroup$ – Piotr Benedysiuk Jul 28 '17 at 19:04
  • $\begingroup$ @PiotrBenedysiuk I did it. If something is unclear, do not hesitate. $\endgroup$ – Duchamp Gérard H. E. Jul 28 '17 at 19:27
1
$\begingroup$

Hint: Note that an eigenfunction is differentiable, than you get an ODE.

(maybe I've a mistake)

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Why is the eigenfunction differentiable with respect to $x$? $\endgroup$ – Piotr Benedysiuk Jul 28 '17 at 18:53
  • $\begingroup$ By the fundamental theorem. $\endgroup$ – I am not Paul Erdos Jul 28 '17 at 18:54
  • $\begingroup$ Lets take $y = \sqrt{x}$. Don't we have $\int f(\sqrt{x}) dx = \int f(y) 2 y dy$? Then, we are using fundamental theorem on a different function - $2xf(x)$ instead of $f(x)$. $\endgroup$ – Piotr Benedysiuk Jul 28 '17 at 18:59
  • $\begingroup$ I don't understand. $ u(x) $ is equal to integral of $ u(\sqrt{s}) $. $\endgroup$ – I am not Paul Erdos Jul 28 '17 at 19:05
  • $\begingroup$ No, the $T(u)(x)$ equals the integral. $\endgroup$ – Piotr Benedysiuk Jul 28 '17 at 19:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.