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I want to bound the following function $y$.

Given a differentiable and monotone decreasing function $x(\tau)$, define $y$ as follows:

$$y(t)= \int_0^t e^{-(t-\tau)} x(\tau) d\tau.$$

While searching about the function, there is no result about boundedness of that function.

Does anybody know?

J Light
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    Take the differentiable and monotonically decreasing $x(\tau ) = \mathrm{e}^{ - \tau /2} $. Then for $t\ge 0$, $$ y(t) = \frac{2}{3}\mathrm{e}^t - \frac{2}{3}\mathrm{e}^{ - t/2} \ge \frac{2}{3}(\mathrm{e}^t - 1) $$ is unbounded. – Gary Dec 27 '22 at 01:33
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    Why the exponential-sum tag? Did you read its description? – metamorphy Dec 27 '22 at 02:47
  • Sorry, I wrote wrong formula and corrected it. – J Light Dec 28 '22 at 00:27
  • I guess there should be a plus sign in brackets? – openspace Dec 28 '22 at 00:28
  • On what domain $x(\tau)$ is assumed to be differentiable? The domain $\tau>0$ is not enough (take $x(\tau)=1/\tau$). If it is continuous for $\tau\ge 0$ and bounded, $|x(\tau)|\leq K$, then $$ \left| {y(t)} \right| \le \int_0^t {{\rm e}^{ - (t - \tau )} \left| {x(\tau )} \right|{\rm d}\tau } \le K\int_0^t {{\rm e}^{ - (t - \tau )} {\rm d}\tau } = K(1 - {\rm e}^{ - t} ) \le K $$ for any $t\ge 0$. – Gary Dec 28 '22 at 00:54

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Why should the function be bounded? If $x(\tau) = 1,$ then $y(t) = e^t(1-e^{-t}) = e^t - 1,$ which does not seem very bounded.

Gary
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Igor Rivin
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