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Let $f$ be a continuous function on $[0,\infty)$ and let $C\geq 0$ be a constant such as $$ \int_x^{x+1} |f(t)|\,\mathrm{d} t\leq C\ \ ,\ \forall x\geq 0. $$ Prove that

$$ e^{-x}\int_0^x e^t|f(t)|\,\mathrm{d} t\leq C\cdot\frac{e}{e-1}\ \ ,\ \forall x\geq 0. $$

PROOF

So far I have tried to seperate the first integral in two integrals and use integration by parts. My next thought was to use absolute value inequalities. I think this method is very wrong. Any help?

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  • $\begingroup$ Is the text at the end of the second integral supposed to mean "for each $x \geq 0$?" $\endgroup$ – Sean Roberson Jul 19 '17 at 19:05
  • $\begingroup$ Yes I'm sorry. Forgot to change the language. $\endgroup$ – mac Jul 19 '17 at 19:12
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Make the substitution $u = x-t$. Your integral expression becomes

$$\int_0^x e^{-u} \lvert f(x-u)\rvert\,du.$$

For $k < \lfloor x\rfloor$, estimate

$$\int_k^{k+1} e^{-u}\lvert f(x-u)\rvert\,du \leqslant e^{-k} \int_k^{k+1} \lvert f(x-u)\rvert\,du \leqslant C e^{-k},$$

and for $k = \lfloor x\rfloor$, estimate

$$\int_k^x e^{-u}\lvert f(x-u)\rvert\,du \leqslant e^{-k} \int_k^x \lvert f(x-u)\rvert\,du \leqslant C e^{-k}.$$

Summing up, we obtain

$$\int_0^x e^{-u} \lvert f(x-u)\rvert\,du \leqslant C\cdot \sum_{k = 0}^{\infty} e^{-k} = C\frac{e}{e-1}.$$

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  • $\begingroup$ A silly question. Sould't the upper limit $x$ in the 1st integral, change as well to $u+t$? $\endgroup$ – mac Jul 19 '17 at 20:26
  • $\begingroup$ No, the bounds become $u = x-0 = x$ for $t = 0$ and $u = x - x = 0$ for $t = x$, and the $dt$ becomes $d(x-u) = -du$, so we get $\int_x^0 e^{-u}\lvert f(x-u)\rvert\, (-du)$, and we use the $-$ from $-du$ to swap the integral limits. $\endgroup$ – Daniel Fischer Jul 19 '17 at 20:36
  • $\begingroup$ So simple but I couldn't see it. Thank you. $\endgroup$ – mac Jul 19 '17 at 20:40

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