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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.

Edit: 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).

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.

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 $\int_{0}^t f(x)^2 d\mu(x)$ in terms of $t$, $m$, and $\mu$ (and nothing else). We currently have the following bound $$\int_{0}^t f(x)^2 d\mu(x) \ge \frac{ m^2 t^4}{36} \mu[0,t].$$ 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.

Edit: 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 $[0,1]$

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.

Edit: 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).

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 $\int_{0}^t f(x)^2 d\mu(x)$ in terms of $t$, $m$, and $\mu$ (and nothing else). We currently have the following bound $$\int_{0}^t f(x)^2 dx \ge \frac{ m^2 t^4}{36} \mu[0,t].$$ 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.

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 $\int_{0}^t f(x)^2 d\mu(x)$ in terms of $t$, $m$, and $\mu$ (and nothing else). We currently have the following bound $$\int_{0}^t f(x)^2 d\mu(x) \ge \frac{ m^2 t^4}{36} \mu[0,t].$$ 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.

Edit: 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 $[0,1]$