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.

Constant Function

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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  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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