Let $f\colon[0, 1] \to \mathbb R$ be an $m$strongly convex function and $\mu$ be a probability measure on $[0,1].$ For any $t<1$, the goal is to find a lower bound on $\int_{0}^t f^2(x) d\mu(x)$ in terms of $t$, $m$, and $\mu$ (and nothing else). We currently have the following bound $$\int_{0}^t f^2(x) d\mu(x) \ge \frac{ m^2 t^4}{36} \mu[0,t].$$ We do not know if our bound is tight. Moreover, our proof is really long and messy. A clean/simple proof of such an elementary result would be helpful.
1 Answer
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The lefthand side, $\int_{0}^t f^2(x)\, dx$, of your inequality does not contain $\mu$. Since you wanted "to find lowerbound on $\int_{0}^t f^2(x) d\mu(x)$", it appears that your desired inequality is $$\int_0^t f^2(x)\,\mu(dx)\ge cm^2t^4 \mu[0,t] \tag{1}$$ for some real $c>0$.
However, (1) is obviously false in general. E.g., let $t=1/2$, $m=1$, $f(x)\equiv x^2/2$, and let $\mu$ be the Dirac measure supported on $\{0\}$. Then $f$ is $m$strongly convex and $\int_0^t f^2(x)\,\mu(dx)=0$, so that (1) fails to hold for any $c>0$.

$\begingroup$ Sorry about that @Iosif. An assumption I forgot to mention, we assume that $\mu$ has a density with respect to the Lebesgue measure, and the density is bounded on $[0,1]$ (but not bounded away from zero) Thank you for your answer and time, however. $\endgroup$– StatguyCommented Jun 27, 2020 at 15:07

$\begingroup$ @Statguy : Your question was fully answered. If you actually wanted to ask something else, you can do that in a separate post. $\endgroup$ Commented Jun 28, 2020 at 2:21