Exercise 8.11

Question

Exercise 11: 
    Suppose $f\in\mathscr R$ on $[0,A]$ for all $A<\infty$, and $f(x)\rightarrow 1$ as $x\rightarrow+\infty$. Prove that $$\lim_{t\rightarrow 0}\int_0^\infty e^{-tx}f(x)\,dx=1\quad\quad(t>0).$$

analambanomenos · Accepted Answer

\def\eps{\varepsilon} Since $|e^{-tx}|\le 1$ for $x\ge 0$ and $t>0$, we have, for $A>0$ and $t>0$ $$\lim_{t\rightarrow0}\Bigg|t\int_0^Ae^{-tx}f(x)\,dx\Bigg|\le\lim_{t\rightarrow0}t\int_0^A\big|f(x)\big|\,dx=0. \leqno{(*)}$$ Let $\varepsilon>0$ and let $A$ be large enough so that $1-\varepsilon0$, $$t\int_A^BKe^{-tx}\,dx= K(e^{-tA}-e^{-tB})\rightarrow Ke^{-tA}\hbox{ as }B\rightarrow\infty.$$ Hence $$(1-\varepsilon)e^{-tA}=t\int_A^\infty(1-\varepsilon)e^{-tx}\,dx\le t\int_A^\infty e^{-tx}f(x)\,dx\le t\int_A^\infty(1+\varepsilon)e^{-tx}\,dx=(1+\varepsilon)e^{-tA}.$$ Letting $t\rightarrow 0+$, we get $$(1-\varepsilon)\le\lim_{t\rightarrow0}t\int_A^\infty e^{-tx}f(x)\,dx\le(1+\varepsilon).$$ Combining this with $(*)$, we get $$(1-\varepsilon)\le\lim_{t\rightarrow0}t\int_0^\infty e^{-tx}f(x)\,dx\le(1+\varepsilon).$$ Since $\varepsilon$ was arbitrary, we finally get $$\lim_{t\rightarrow0}t\int_0^\infty e^{-tx}f(x)\,dx=1.$$

Exercise 8.11

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Exercise 8.11

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