# Lower bound on $L^2$ norm of a strongly convex function

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.

The left-hand 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 $$c>0$$.
• 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. Commented Jun 27, 2020 at 15:07