# Constant Function

Let $f : \mathbb{R} \to \mathbb{R}$ be a continous and bounded function such that $f(x) = \int_x^{x+1} f(y) dy$ for all $x \in \mathbb{R}$

Then $f$ is constant.

## 5 thoughts on “Constant Function”

1. Because $f$ is continuous and integrals of continuous functions are differentiable, it is evident that $f$ is smooth, and $f'(x) = f(x+1)-f(x)$ for every $x\in\mathbb R$. In fact, $f^{(n)}(x) = \Delta^n f(x)$ for every $n>0$, so that if $M$ is a bound for $f$, $2^n M$ is a bound for $f^{(n)}(x)$. This proves that $f$ must be analytic, and that in fact that $f(X) = e^{-X} \sum_{k\geqslant 0} f(k) X^k/k!$. But then $f$ defines an holomorphic function on all of $\mathbb C$ and [TODO].

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2. Just an aside on whether the boundedness is needed or not: as noted above, any $f$ as in the statement satisfies $f'(x)=f(x+1)-f(x)$, which is a linear delay differential equation (DDE) with constant coefficients and discrete delays with associated characteristic equation $e^\lambda =\lambda+1$ (see for instance section 1.3 of this thesis). A bound for $f$ may probably be obtained from this.

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1. Obviously there are non-bounded solutions, like $f(x)=ax+b$, but they do not satisfy the original integral equation…

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3. nachodarago says:

Recall that the Laplace transform of a continuous bounded function $f:(0,+\infty)\to\mathbb{R}$ is defined as $\mathcal{L}f(s) = \displaystyle\int_{0}^{\infty} e^{-st}f(t)dt$, and this defines an operator $\mathcal{L}:\mathcal{C}_b(0,\infty)\to \mathcal{C}_b(0,\infty)$ where $\mathcal{C}_b(0,\infty)$ is the space of continuous bounded functions over the positive reals. Furthermore, this operator is an isomorphism of Banach spaces.

Integrating by parts, it is easy to see that if $f$ is a continuously differentiable function such that $f'$ is bounded, then $\mathcal{L}f'(s) = s\mathcal{L}f(s)-f(0)$. It is also easy to see that if $g(t) = f(t+1)$ then $\mathcal{L}g(s) = e^s\mathcal{L}f(s)$.

Now, let $f$ be a function that satisfies the hypothesis. WLOG we can assume that $f(0)=0$. Since the hypothesis implies that $f'(s)=f(s+1)-f(s)$ for all $s>0$, by applying the Laplace transform we obtain $(s-e^s+1)\mathcal{L}f(s)=0$ for all $s>0$, and thus $\mathcal{L}f(s)=0$. Since it is an isomorphism for bounded functions, we get $f(s)=0$ for $s>0$. By noticing that $g(s)=f(-s)$ satisfies the hypothesis we obtain $f(s)=0$ for all $s<0$. And we are done.

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