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Integral representation of bounded and absolutely integrable functions. (English) Zbl 0957.26001
The aim of this paper is to obtain an integral representation formula for an even function.
First of all, the author shows that if \(f(x)\) is an even function over the interval \((-1,1)\), \(G(u)\) is an even bounded integrable function over \((-1,1)\), \(G(u)\) together with its derivatives of all orders is continuous over \((-1,1)\) and it vanishes with all its derivatives for \(u=F(1)\), then for \(\int^1_{-1}G(u)du\neq 0\), the representation formula \[ f(x)=\lim_{n\to\infty} \frac 1{n!\int^1_{-1}G(u) du} \int^1_{-1}\frac{d^{n+1}} {du^{n+1}}\left[(u-x)^n\int^u_{-1}G(\overline u)d\overline u\right]f(u) du \] holds.
The main result states that if \(f(x)\) is bounded and absolutely integrable over \((0,1-\varepsilon)\), is zero over \((1-\varepsilon,1)\) and satisfies \[ f(x)=\lim_{n\to\infty} \frac 1{n!F(1)}\int^1_0\frac{d^{n+1}}{du^{n+1}} \left\{[(u-x)^n+(u+x)^n] F(u)\right\}f(u) du \tag{1} \] at every point of \((0,1)\), then it is identically zero over \((0,1)\), where \(F(u):=\int^u_{-1}G(\widetilde u) d \widetilde u\).
Hence, a function satisfying (1) over \((0,1)\), bounded and absolutely integrable, is completely characterized by its value in the neighborhood of 1. Instead of (1) it is enough to suppose that the following limit exists \[ \lim_{n\to\infty} \frac 2{n!F(1)} \int^1_0 \frac{d^{n+1}}{du^{n+1}}[u^n F(u)] f(u)du . \]
26A42 Integrals of Riemann, Stieltjes and Lebesgue type
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