As presented in the question's title, I wish to find a function $\lambda(\cdot): [0, \infty) \to [0, \infty)$ which satisfies the integral equation:

\begin{equation} \frac{1}{\lambda(y)} = c_1 \int_0^\infty \lambda(x) \exp(-c_2 y x) \, dx \end{equation}

where $c_1$ and $c_2$ are positive constants.

Unfortunately, I have no clear idea about how to systematically tackle this question. Any help is greatly appreciated!

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    $\begingroup$ You can easily get rid of $c_1$ and $c_2$. Just an idea. $\endgroup$ – iamvegan Oct 13 '16 at 14:30
  • $\begingroup$ I agree about getting rid of $c_1$. As far as $c_2$ is concerned, I don't see a way; for instance any substitution involving $c_2 x$ inevitably places $c_2$ into the argument of the integrated $\lambda$. $\endgroup$ – R.G. Oct 13 '16 at 14:35

Let me show that no such function exists. We argue this by contradiction. Assume that there is a function $\lambda : [0, \infty) \to [0, \infty)$ satisfying

$$ \frac{1}{\lambda(s)} = c_1 \int_{0}^{\infty} \lambda(x) e^{-c_2 sx} \, dx \quad \forall s \geq 0 \tag{*}$$

with the convention that $1/0 = \infty$. Then

Step 1. In this step, we normalize $\lambda$ and reveal some useful facts on it.

  • Since the right-hand side of $\text{(*)}$ is decreasing, $\lambda$ is increasing.

  • Since the left-hand side of $\text{(*)}$ is always positive (or possibly infinite), $\lambda$ cannot be identically zero.

  • From these two properties, $\alpha$ defined by $$ \alpha := \inf \{ s > 0 : \lambda(s) > 0 \} $$ Is a non-zero real number such that $\lambda(s) = 0$ for all $s \in [0, \alpha)$. Moreover, if $\alpha > 0$ then the modified version $\tilde{\lambda}(s) = e^{-c_2 \alpha s}\lambda(s+\alpha)$ satisfies \begin{align*} \frac{1}{\tilde{\lambda}(s)} = \frac{e^{c_2 \alpha s}}{\lambda(s+\alpha)} &= c_1 e^{c_2 \alpha s} \int_{\alpha}^{\infty} \lambda(x) e^{-c_2 (s+\alpha) x} \, dx \\ &= c_1 e^{c_2 \alpha s} \int_{0}^{\infty} \lambda(x+\alpha) e^{-c_2 (s+\alpha)(x+\alpha)} \, dx \\ &= c_1 e^{-c_2 \alpha^2} \int_{0}^{\infty} \tilde{\lambda}(x) e^{-c_2 sx} \, dx. \end{align*} So by changing the value of $c_1$ if needed, we may assume that $\alpha = 0$. Then by @guestDiego's computation, we may further assume that $c_1 = c_2 = 1$ and we do so.

  • If $\lambda(0) > 0$, then $$ \infty > \frac{1}{\lambda(0)} = \int_{0}^{\infty} \lambda(x) \, dx \geq \int_{0}^{\infty} \lambda(0) \, dx = \infty $$ and we get a contradiction. Thus $\lambda(0) = 0$.

  • From the standard theory of Laplace transform, if $f \geq 0$ is measurable and $$\mathcal{L}\{f\}(s) := \int_{0}^{\infty} f(x)e^{-sx} \, dx$$ Is finite for all $s > 0$, then $\mathcal{L}\{f\}(s)$ converges for $\Re(s) > 0$ and defines an analytic function on the same region. Moreover, differentiation can be computed by using Leibniz's integral rule: $$ \frac{d^n}{ds^n} \mathcal{L}\{f\}(s) = (-1)^n \int_{0}^{\infty} x^n f(x) e^{-sx} \, dx. $$

Step 2. Now we are ready to establish a contradiction. First, we have $\lambda'(s) \geq 0$ because $\lambda$ is increasing. Then by the Tonelli's theorem,

\begin{align*} \frac{s}{\lambda(s)} &= \int_{0}^{\infty} \lambda(t) s e^{-st} \, dt \\ &= \int_{0}^{\infty} \bigg( \int_{0}^{t} \lambda'(x) \, dx \bigg) s e^{-st} \, dt \\ &= \int_{0}^{\infty} \lambda'(x) \bigg( \int_{x}^{\infty} s e^{-st} \, dt \bigg) dx \\ &= \int_{0}^{\infty} \lambda'(x) e^{-sx} \, dx. \end{align*}

Taking log-differentiation to both sides, we get

$$ \frac{\lambda'(s)}{\lambda(s)} = \frac{\int_{0}^{\infty} x \lambda'(x) e^{-sx} \, dx}{\int_{0}^{\infty} \lambda'(x) e^{-sx} \, dx} + \frac{1}{s}. \tag{1}$$

This is our key ingredient toward a contradiction. Using this, we inductively prove that

Claim. For any $n = 1, 2, 3, \cdots$ and $s > 0$, we have $\frac{\lambda'(s)}{\lambda(s)} \geq \frac{n}{s}$.

The base case $n = 1$ is straightforward from $\text{(1)}$ since the ratio between two integra in the RHS of $\text{(1)}$ is non-negative. Next, assuming the claim for $n$, we have

$$ \frac{\lambda'(s)}{\lambda(s)} \geq \frac{\int_{0}^{\infty} n \lambda(x) e^{-sx} \, dx}{\int_{0}^{\infty} \lambda'(x) e^{-sx} \, dx} + \frac{1}{s} = \frac{n/\lambda(s)}{s/\lambda(s)} + \frac{1}{s} = \frac{n+1}{s}. $$

Then the claim follows from mathematical induction.

Now the contradiction is obvious: $\lambda'(s)/\lambda(s)$ is finite for $s > 0$ while $n/s$ can be arbitrary large! Therefore no such $\lambda$ can exist.

  • $\begingroup$ Very nice argument! May I ask what the intuition for showing $\frac{s}{\lambda(s)} = \int_0^\infty \lambda'(x) \exp(-s*x) \, dx$ is? $\endgroup$ – R.G. Oct 18 '16 at 12:37
  • $\begingroup$ @R.G., it is essentially integration by parts. But you can avoid introducing some nebulous intermediate term if you use Fubini's theorem. $\endgroup$ – Sangchul Lee Oct 18 '16 at 13:03
  • $\begingroup$ I'm sorry for the being unclear with my question. It is not the calculation I want to know about but rather why you thought that evaluating this will help you later on. $\endgroup$ – R.G. Oct 18 '16 at 13:13
  • $\begingroup$ @R.G. Now I see what you asked. I had no clear intuition, but somehow I believed that higher derivatives of $\lambda$ would reveal erratic behavior which we may employ to derive a contradiction. (To be precise, I believed that $\lambda^{(n)}$ will change sign for some $n$ so that the Laplace transform of it becomes negative, which is impossible in view of $s^n / \lambda(s) = \mathcal{L}\{\lambda^{(n)}\}(s)$.) So it was natural for me to consider Laplace transform of derivatives of $\lambda$. It was totally unclear to me where it will lead me, but very fortunately $\text{(1)}$ was enough! $\endgroup$ – Sangchul Lee Oct 18 '16 at 13:21
  • $\begingroup$ Thanks for this insight! I hope to reach a similar level of mathematical maturity one day. $\endgroup$ – R.G. Oct 19 '16 at 7:51

I make explicit the hint of iamavegan.

If $\lambda_1(y)=\sqrt{c_1}\lambda(y)$, then $$ \frac{1}{\lambda_1(y)} = \int_0^\infty \lambda_1(x) \exp(-c_2 y x) \, dx= \int_0^\infty \frac{\lambda_1(s/\sqrt{c_2})}{\sqrt{c_2}} \exp(-\sqrt{c_2} y s) \, ds $$ Define $$ \lambda_2(z)=\frac{\lambda_1(z/\sqrt{c_2})}{c_2^{1/4}}. $$ Then, with $z=y\sqrt{c_2} $, $$ \frac{1}{\lambda_2(z)} = \int_0^\infty \lambda_2(s) \exp(-sz) \, ds. $$ and $$ \lambda(y)=c_1^{-1/2}c_2^{1/4}\lambda_2(y c_2^{1/2}) $$

  • 1
    $\begingroup$ Thank you very much for this elaboration! $\endgroup$ – R.G. Oct 13 '16 at 15:08

HINT: iterate the $\lambda$ into the integral. \begin{align*} \frac1{\lambda(y)} &=c_1\int_0^{+\infty}\lambda(x)e^{-c_2xy}\,dx\\ &=c_1\int_0^{+\infty}\frac{e^{-c_2xy}}{c_1\int_0^{+\infty}\lambda(z)e^{-c_2zx}\,dz\\}\,dx\\ &=\int_0^{+\infty}\frac{e^{-c_2xy}}{\int_0^{+\infty}\lambda(z)e^{-c_2zx}\,dz\\}\,dx\\ \end{align*} go on from here

  • $\begingroup$ Thank you for this hint. I'll see where this will take me. $\endgroup$ – R.G. Oct 13 '16 at 15:09

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