# a problem in functional analysis or operator theory

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.

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

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 $U(u)(x)=u(x^2)$ ($T=UWU^{-1}$).

• First off, thanks. $F'(x^{2})=2xu(x)$ then $u= e^{\frac{4x^{3}}{3\lambda}}.$ Any way, I could not understand the contradiction. – Hamit Jul 29 '17 at 7:09
• $F'(X)=u(\sqrt{X})$ 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. – Duchamp Gérard H. E. Jul 29 '17 at 7:20
• @HamedBaghalGhaffari Thanks. Do not hesitate however if something remains unclear. – Duchamp Gérard H. E. Jul 30 '17 at 14:19
• Thank you. It seems that I have to practice more in order to be accustomed to solve math problems. – Hamit Jul 31 '17 at 9:59
• @HamedBaghalGhaffari You're right, the successful recipe being : practice + interaction. Good luck ! – Duchamp Gérard H. E. Jul 31 '17 at 13:31

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.

• [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 ? – Duchamp Gérard H. E. Jul 30 '17 at 4:05
• 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 :) – Duchamp Gérard H. E. Jul 30 '17 at 4:59
• @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. – Ranc Jul 30 '17 at 6:32
• @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 $\lambda^{-n}$). – Ranc Jul 30 '17 at 6:37
• @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. – Ranc Jul 30 '17 at 6:47

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 $x=0$

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

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.

• $\frac{2}{3} x^{3}=\lambda x$ so if $\lambda =1,$ then $u(x)=x$ is an answer for the question which is a contradiction. – Hamit Jul 28 '17 at 18:49
• @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 $x=1$. – Piotr Benedysiuk Jul 28 '17 at 18:52
• Good point. Thank you. I did not pay attention to the domain. Now, I have to find a proof for the problem... – Hamit Jul 28 '17 at 18:57
• @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. – Piotr Benedysiuk Jul 28 '17 at 19:04
• @PiotrBenedysiuk I did it. If something is unclear, do not hesitate. – Duchamp Gérard H. E. Jul 28 '17 at 19:27

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

(maybe I've a mistake)

• Why is the eigenfunction differentiable with respect to $x$? – Piotr Benedysiuk Jul 28 '17 at 18:53
• By the fundamental theorem. – I am not Paul Erdos Jul 28 '17 at 18:54
• 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 $f(x)$. – Piotr Benedysiuk Jul 28 '17 at 18:59
• I don't understand. $u(x)$ is equal to integral of $u(\sqrt{s})$. – I am not Paul Erdos Jul 28 '17 at 19:05
• No, the $T(u)(x)$ equals the integral. – Piotr Benedysiuk Jul 28 '17 at 19:07