Assume that we have:
$g(t)+\int_{t}^{T}f(s)ds\leq h(t)$ for all $0\leq t\leq T$
with $f$ is a positive increasing function. Can me deduce or write
$g(t)+\int_{0}^{T}f(s)ds\leq h(t)$ for all $0\leq t\leq T$ ?
Integral function and inequality
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integration
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0When ,we get $∫_{T/2}^{T }f(s)ds$$\leq $$∫_{t}^{T}f(s)ds$? – 2017-01-11
1 Answers
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No. To see that it does not work in general, if $$ h(t)=g(t)+\int_t^T f(s)\,ds $$ then the first inequality is satisfied, but, since $f\geq 0$, $$ g(t)+\int_0^T f(s)\,ds > g(t)+\int_t^T f(s)\,ds=h(t) $$ if $t>0$.