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$f'(x)=f(x+1)-f(x)$, $f(x)\geq 0$, $f$ is continuous on $\Bbb R$, then $f$ is constant

I know many topics considering the following problem without $f\geq 0$. It or its method does not fit it. Can you help?

$f'(x)+f(x)=f(x+1)\geq 0$, $e^xf(x)$ is increasing....then?

xldd
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2 Answers2

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Suppose $f : \mathbb{R} \to \mathbb{R}$ is a differentiable function such that

$$ \forall x \in \mathbb{R} \quad : \quad f(x) \geq 0 \quad \text{and} \quad f'(x) = f(x+1) - f(x). $$

Now set $g(x) =e^x f(x)$. Then it is easy to verify that $g \geq 0$ and

$$ g'(x) = e^{-1} g(x+1), \quad \text{i.e.,} \quad g(x) = e g'(x-1). $$

From this, we know that $g$ is absolutely monotone on $\mathbb{R}$. So by the Bernstein's theorem, there exists a finite Borel measure $\mu$ on $[0, \infty)$ such that

$$ g(x) = \int_{[0, \infty)} e^{xt} \, \mu(\mathrm{d}t), \quad \forall x \leq 0. $$

Now let $x$ be arbitrary, and let $n \in \mathbb{N}$ be any positive integer such that $x \leq n$. Then

\begin{align*} g(x) &= e^n g^{(n)}(x - n) \\ &= e^n \int_{[0, \infty)} t^n e^{(x-n)t} \, \mu(\mathrm{d}t) \\ &= \int_{[0, \infty)} e^{xt} (t e^{1-t})^n \, \mu(\mathrm{d}t). \end{align*}

However, for $t \geq 0$, we have $t e^{1-t} \leq 1$ with the equality if and only if $t = 1$. So, if

$$ h_{n}(t) = e^{xt} (t e^{1-t})^n $$

denotes the the integrand, then we observe:

  1. $h_{n}(t)$ is non-negative and monotone decreasing in $n$,
  2. $h_{n}(t)$ is $\mu$-integrable for any sufficiently large $n$, and
  3. $h_{n}(t)$ converges pointwise to $e^x \mathbf{1}_{\{1\}}(t)$ as $n \to \infty$.

So by letting $n \to \infty$, the dominated convergence theorem shows that

\begin{align*} g(x) &= \int_{[0, \infty)} e^{xt} \mathbf{1}_{\{1\}}(t) \, \mu(\mathrm{d}t) = C e^x \end{align*}

for $C = \mu(\{1\})$. Therefore $f(x) = e^{-x}g(x) = C$ is constant.

Addendum - Some Ideas for Another Proof. Let $f$ solve the functional equation. We can prove that, for the compensated Poisson process $X_t = N_t - t$, where $N_t$ is the Poisson process with unit rate, we have

$$ f(a) = \mathbf{E}[f(a+X_t)]. $$

In other words, $f$ is $X$-harmonic, and $f(a+X_t)$ is a martingale.

  • This has to do with the fact that $e^D - 1$ is the infinitesimal generator of $N_t$, where $D = \frac{\mathrm{d}}{\mathrm{d}x}$.

  • Note that the functional equation reads as $e^D f = (1 + D) f$. Now, this yields yet another formal identity $f = e^{t(e^D - 1)} e^{-tD} f$, which translates to the above identity.

I am curious whether we can utilize this fact to give yet another proof that $f$ is constant.

Sangchul Lee
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Partial solution: As you say, the function $e^xf(x)$ is incrreasing, so $$\lim_{x\to\infty}e^xf(x)=A<\infty\quad \text{or}\quad \infty.$$ If $\lim_{x\to\infty}e^xf(x)=A$, then $$\lim_{x\to\infty}f(x)=\lim_{x\to\infty}\frac{e^xf(x)}{e^x}=0.$$ Hence, $$\lim_{x\to\infty}f'(x)=\lim_{x\to\infty}[f(x+1)-f(x)]=0,$$ then by the result in the post $f(x+1)-f(x)=f'(x)$: prove $f(x)$ linear function, we get that $f(x)=ax+b$. Combining $f(x)\to0$, we know that $a=b=0,f(x)\equiv0.$

If $\lim_{x\to\infty}e^xf(x)=\infty$, I have no idea now! Hope someone can give a complete solution!

Riemann
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