$T$ has no eigen-values

Question

I would like to solve the following exercise:

Let $V$ be the space of continuous real-valued functions on the real line. Let $T$ be the linear map on $V$ defined by $$ (Tf)(x) = \int\limits_{0}^{x}f(t)dt.$$ Prove that $T$ has no eigen-values.

Here is my attempt:

Suppose on contrary, $T$ has an eigenvalue say $\lambda$. Then there is a non-zero $f\in V$ such that $$ Tf= \lambda f$$ i.e., for all $x\in \mathbb{R}$, $$ \int\limits_{0}^{x}f(t)dt= \lambda f(x) $$ i.e., for all $x\in \mathbb{R}$, $$ f(x)-f(0)= \lambda f'(x)$$ i.e., for all $x\in \mathbb{R}$,$$ (f(x)-f(0))^{\lambda}= e^{\mu x}$$ Take $x=0$, we get $$ 0=e^{0}=1.$$ Therefore $T$ has no eigenvalues.

Is the above argument correct?

Answer 1

Your argument fails at the point $$f(x) - f(0) = \lambda f'(x), \forall x \quad\implies\quad (f(x)-f(0))^{\lambda} = e^{\mu x},\forall x \tag{*1}$$

Let's say $f_0(x)$ is a solution satisfy LHS, then every constant multiple of $f_0(x)$ is also a solution. However, the RHS of your argument doesn't maintain the same form.

It is clear any eigenfunction $f(x)$ of the problem satisfy the ODE on LHS of $(*1)$. Furthermore, $f(0) = 0$. By Picard–Lindelöf theorem (aka, the uniqueness of solutions of first-order ODE with given initial conditions), we can deduce $f(x) = 0$ identically. This will contradict with the fact that $f(x)$ is an eigenfunction and cannot be zero identically.

Following is a direct proof that $f \mapsto Tf$ doesn't have any eigenvalue. It illustrate the Picard iteration process which appears in the proof of uniqueness part of Picard-Lindelof theorem.

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Let's say $\lambda$ is an eigenvalue of the map $f \mapsto Tf$:

$$Tf(x) = \int_0^x f(x) dx = \lambda f(x),\quad \forall x \ge 0$$

and $f(x)$ is corresponding eigenfunction. Since $f(x)$ is an eigenfunction, it cannot be zero identically. Since $f(x)$ is continuous, we can find a $L > 0$ such that $f(L) \ne 0$.

First, let us consider the case $\lambda = 0$. When $\lambda = 0$, we have

$$\int_0^x f(x) dx = 0, \forall x \in [0,L] \quad\implies\quad f(L) = \left.\frac{d}{dx}\left(\int_0^x f(x) dx \right)\right|_{x=L} = \left.\frac{d}{dx} 0\right|_{x=L} = 0$$ This contradicts with our choice of $L$.

Next, consider the case $\lambda \ne 0$. Since $f(x)$ is continuous over $[0,L]$, we can pick a $M > 0$ such that $|f(x)| \le M$ over $[0,L]$. For any $x \in [0,L]$, above bound leads to

$$|f(x)| = \left|\frac{1}{\lambda}\int_0^x f(x)dx\right| \le \frac{1}{|\lambda|}\int_0^x |f(x)|dx \le M \frac{x}{|\lambda|}$$ this new bound in turn implies

$$|f(x)| \le \frac{1}{|\lambda|}\int_0^1 M\frac{x}{|\lambda|} dx = \frac{M}{2!}\left(\frac{x}{|\lambda|}\right)^2$$ Repeat this procedure $n-2$ more times, we find for any integer $n \ge 1$,

$$|f(x)| \le \frac{M}{n!}\left(\frac{x}{|\lambda|}\right)^n,\quad\forall x \in [0,L]$$

Restrict $x$ to $L$ and sending $n$ to infinity, we obtain

$$|f(L)| \le \lim_{n\to\infty} \frac{M}{n!}\left(\frac{L}{|\lambda|}\right)^n = 0 \quad\implies\quad f(L) = 0$$ This again contradicts with our choice of $L$.

Combine these, we can conclude the map $f \mapsto Tf$ doesn't have any eigenvalue.

Answer 2

The OP Amitks' argument seems deeply flawed to me. My chief difficulty at this point lies in his attempt to infer

$f(x) - f(0) = \lambda f'(x) \tag 1$

from

$Tf(x) = \displaystyle \int_0^x f(t) \; dt = \lambda f(x); \tag 2$

if we differentiate the equation

$\displaystyle \int_0^x f(t) \; dt = \lambda f(x) \tag 3$

we should obtain, if I am not mistaken,

$\lambda f'(x) = f(x); \tag 4$

the term "$-f(0)$" should not occur on the left of (1). After that, our OP asserts that

$(f(x) - f(0))^\lambda = e^{\mu x}; \tag 5$

not to put too fine a point on it but, frankly, I have no idea how (5) follows from (1).

I'll leave off further commentary on Amitks' attempt, turning instead to my analysis of this problem:

We recall to begin that eigenvectors, by definition, cannot vanish; we shall use this fact more than once it what follows.

We also need to know that, for $f(x)$ continuous and fixed $x_0 \in \Bbb R$, $\int_{x_0}^x f(t) \; dt$ is differentiable and

$\left ( \displaystyle \int_{x_0}^x f(t) \; dt \right )' = f(x); \tag 6$

(6) is in fact a statement of the Fundamental Theorem of Calculus, and will find more than one application below.

First let us consider the case $\lambda = 0$; then the eigen-equation $Tf(x) = \lambda f(x)$ reads

$Tf(x) = \displaystyle \int_0^x f(t) \; dt = 0, \; \forall x \in \Bbb R; \tag 7$

if we differentiate (7) and use (6) we find

$ f(x) = \left ( \displaystyle \int_0^x f(t) \; dt \right )' = 0, \; \forall x \in \Bbb R; \tag 8$

since eigenfunctions cannot identically vanish, this shows that $0$ cannot be an eigenvalue of $T$.

Now for $\lambda \ne 0$, the eigen-equation appears as in (2); we may as we have said differentiate this to yield

$\lambda f'(x) = f(x), \; \forall x \in \Bbb R, \tag 9$

or

$f'(x) = \lambda^{-1} f(x); \tag{10}$

at this point it proves most convenient to invoke a little bit from the elementary theory of ordinary differential equations and assert the uniqueness of the solution to (10) for any $f(0)$; that is,

$f(x) = f(0) e^{\lambda^{-1}x} \tag{11}$

is the only function satisfying (10) for a given value of $f(0)$; then

$Tf(x) = \displaystyle \int_0^x f(t) \; dt = f(0) \int_0^x e^{\lambda^{-1} t} \; dt = f(0) \lambda ( e^{\lambda^{-1} x} - e^0 ) = f(0) \lambda (e^{\lambda^{-1} x} - 1), \tag{12}$

but

$\lambda f(x) = f(0) \lambda e^{\lambda^{-1}x}, \tag{13}$

whence

$Tf(x) = f(0) \lambda (e^{\lambda^{-1} x} - 1) \ne f(0) \lambda e^{\lambda^{-1} x} = \lambda f(x) \tag{14}$

unless $f(0) = 0$; but then by (11) this forces

$f(x) = 0, \; \forall x \in \Bbb R, \tag{15}$

which is forbidden according to the definition of eigenvector. So $\lambda \ne 0$ cannot be an eigenvalue of $T$.

We conclude that $T$ has no eigenvalues whatsoever.