Verify whether the following claim is true or false :
Let $f$ be a continuous and nonincreasing function on $[0,+\infty)$ such that $$\lim_{t\to+\infty}\int_{0}^{t}(f(s))^{2}ds<+\infty.$$ Then , $$\lim_{t\to +\infty}f(t)=0$$
Since a possible counterexample didn't draw my attention , my approach so far :
Let $\displaystyle F(t)=\int_0^t(f(s))^2ds$ and assume the contrary (yes , I assumed prior existence of that limit) that $\displaystyle\lim_{t\to+\infty}f(t)=l\neq0$ . Then by FTC $$F'(t)=(f(t))^2\to l^2>0 \ \text{as} \ t\to+\infty$$ Thus $F(t)$ is strictly increasing as $t\to+\infty$ . Afterwards I was unable to get a contradiction with $\displaystyle \lim_{t\to+\infty}F(t)<+\infty$ . Any help is appreciated in this regard .
If $\lim_{t\to+\infty}f(t)=l\not=0$ then there is $R>0$ such that $f^2(t)>\frac{l^2}{2}$ for any $t>R$. Therefore $$\int_{0}^{t}(f(s))^{2}\, ds>\int_{0}^{R}(f(s))^{2}ds +\int_{R}^{t}\frac{l^2}{2} \,ds=\int_{0}^{R}(f(s))^{2}ds +\frac{l^2}{2}(t-R)\to +\infty$$ as $t\to +\infty$.