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In many sources, for example here, we can find the following generalisation of Jensen's inequality.

In real analysis, we may require an estimate on $\varphi \left(\int_{a}^{b}f(x)\,dx\right)$, where ${a,b\in \mathbb {R} }$ and $f:[a,b]\to \mathbb {R}$ is a non-negative Lebesgue-integrable function. In this case, the Lebesgue measure of ${[a,b]}$ need not be unity. However, by integration by substitution, the interval can be rescaled so that it has measure unity. Then Jensen's inequality can be applied to get ${\varphi \left({\frac {1}{b-a}}\int_{a}^{b}f(x)\,dx\right)\leq {\frac {1}{b-a}}\int _{a}^{b}\varphi (f(x))\,dx.}$

Can we generalise this statement to indefinite integrals? In particular, is the following inequality true: $\left(\int_{-\infty}^t f(x) dx\right)^2 \le \int_{-\infty}^t \left(f(x)\right)^2 dx$?

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No, this is not true. Take, for instance, $f(x)=e^x$. Then $$\int_{-\infty}^t (f(x))^2 \ dx = \int_{-\infty}^t e^{2x} \ dx = \frac{1}{2}e^{2x}\Bigg|^{x=t}_{x=-\infty} = \frac{1}{2}e^{2t}$$ whereas $$\left(\int_{-\infty}^t f(x) \ dx \right)^2 = \left(\int_{-\infty}^t e^x \ dx \right)^2= \left(e^x \Big|_{x=-\infty}^{x=t}\right)^2 = (e^t)^2 = e^{2t}$$ so $\left(\int_{-\infty}^t f(x) \ dx \right)^2 > \int_{-\infty}^t (f(x))^2 \ dx$.

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  • $\begingroup$ Thank you for providing the counterexample, which obviously proves that we cannot generalise the Jensen't inequality in a way I suggested. $\endgroup$ – Paula Apr 30 '17 at 15:27

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