# Special function or Weierstrass?

Given that $f(x)$ is integrable on any $[0,a]$, such that $\lim_{x\to+\infty} f(x)=1$

Prove that: $$\lim_{t\to 0+}\int_0^{\infty}t\text{e}^{-tx}f(x)\text{d}x=1$$

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What does that title mean? Might you replace it with something less obscure, more informative? – Gerry Myerson Feb 3 '13 at 9:46
@GerryMyerson : It is a question in Rudin's special function chapter...I dont see anything relevant to that chapter :( eventhough I know a proof – Ryan Feb 3 '13 at 9:57
Ryan, that doesn't answer my questions. Also, Rudin wrote several books. – Gerry Myerson Feb 3 '13 at 10:03
@GerryMyerson : His book on Analysis – Ryan Feb 3 '13 at 10:08
@Ryan: Rudin wrote several books about analysis... – Hans Lundmark Feb 3 '13 at 10:40

The function $g:z\mapsto\sup\limits_{x\geqslant z}|f(x)-1|$ is finite everywhere and, for every $z$, $$\left|\int_0^{+\infty}t\mathrm e^{-tx}f(x)\mathrm dx-1\right|\leqslant(1-\mathrm e^{-tz})\cdot g(0)+\mathrm e^{-tz}\cdot g(z)\leqslant t\cdot z\cdot g(0)+ g(z)$$ Thus, for every $z$, $$\limsup\limits_{t\to0}\ \left|\int_0^{+\infty}t\mathrm e^{-tx}f(x)\mathrm dx-1\right|\leqslant g(z).$$ Since $g$ has limit $0$ at $+\infty$, the proof is complete.