Let $0<\alpha<1/2$, $\lambda\ge1$ and $t\in[0,T]$, where $T>0$ is fixed; how can I prove that $$ \int_0^{\lambda t}(\lambda t-y)^{-\alpha}\,dy\le\sup_{z>0}\int_0^ze^{-y}(z-y)^{-\alpha}\,dy\le4\;\;\;\;? $$ The integral on RHS, without the $\sup$ would be smaller than the one on LHS; why can the $\sup$ make this inequality come true? And how can we obtain the numerical bound given by $4$?
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What do you know about $t$? – Umberto P. Mar 27 '17 at 19:05
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@UmbertoP. sorry you're right! I've edited! – Joe Mar 27 '17 at 19:07
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Any relationship between $T$, $\alpha$, and $\lambda$? – Umberto P. Mar 27 '17 at 19:10
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The integral on the left can be evaluated explicitly: $$\int_0^{\lambda t} ( \lambda t - y)^{-\alpha} \, dy = \left.- \frac{(\lambda t - y)^{-\alpha + 1}}{-\alpha + 1} \right|_{y = 0}^{y = \lambda t} = \frac{(\lambda t)^{1 - \alpha}}{1 - \alpha}.$$
Unless you can control how large $t$ is permitted to get this can be arbitrarily large.
Umberto P.
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I know this, in fact $t\in[0,T]$ , where $T>0$ is FIXED, thus $t$ can't be arbitrarily large; I wrote this in my edit. – Joe Mar 27 '17 at 19:13
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What if it is FIXED so that $\dfrac{(\lambda T)^{1 - \alpha}}{1 - \alpha} = 5$? – Umberto P. Mar 27 '17 at 19:14
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The bound with $4$ should be proved with the other integral, the one with the supremum, not with the first one. One once we do this, how can we prove the inequality between the two integrals? – Joe Mar 27 '17 at 19:16
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1You can choose $\lambda$, $\alpha$, and $T$ to make the integral on the left equal to $5$. One of these inequalities isn't true. – Umberto P. Mar 27 '17 at 19:27
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You are right. My initial inequality was wider; I'm gonna open another post immediately. However thanks. – Joe Mar 27 '17 at 19:31