Assume $f : [0,\frac{\pi}{2}] \to (0,+\infty)$ is continuous, and for $\epsilon>0$ small enough, there is $$ \left(\frac{1}{\pi/2}\int_0^{\pi/2} \sqrt{f(t)}dt\right)^2 > \epsilon+ \int_0^{\pi/2} f(t)\sin(t)dt $$ Then how to show that for any continuous function $g : [0,\frac{\pi}{2}] \to (0,+\infty)$ satisfy $$ \int_0^{\pi/2} |f(t)-g(t)|dt \le \frac{1}{10}\epsilon $$ there is $$ \left(\frac{1}{\pi/2}\int_0^{\pi/2} \sqrt{g(t)}dt\right)^2 > \frac{2}{3}\epsilon+ \int_0^{\pi/2} g(t)\sin(t)dt ~~~? $$
This problem is from there, I feel should use Sobolev embedding, but I fail.
I don't know how to make the proof work with $\frac{\epsilon}{10}$, since in my proof, I would need $(a+b)^2\le a^2+b^2$, which is false. But since you are using density, you can take any small quantity you want. Fix $0<\varepsilon<\epsilon$ and take $g$ so that $$ \int_{0}^{\pi/2}|f(t)-g(t)|\,dt\leq\varepsilon $$ You have \begin{align*} \int_{0}^{\pi/2}g(t)\sin t\,dt & =\int_{0}^{\pi/2}f(t)\sin t\,dt+\int% _{0}^{\pi/2}(g(t)-f(t))\sin t\,dt\\ & \leq\int_{0}^{\pi/2}f(t)\sin t\,dt+\int_{0}^{\pi/2}|f(t)-g(t)|\,dt\\ & \leq\int_{0}^{\pi/2}f(t)\sin t\,dt+\varepsilon\leq\int_{0}^{\pi/2}f(t)\sin t\,dt+\frac{\epsilon}{3}% \end{align*} provided $\varepsilon\leq\frac{\epsilon}{3}$. On the other hand, by Holder's inequality \begin{align*} \int_{0}^{\pi/2}\sqrt{f(t)}\,dt & =\int_{0}^{\pi/2}\sqrt{g(t)+(f(t)-g(t))}% \,dt\\ & \leq\int_{0}^{\pi/2}\sqrt{g(t)}\,dt+\int_{0}^{\pi/2}\sqrt{|f(t)-g(t)|}% \,dt\\ & \leq\int_{0}^{\pi/2}\sqrt{g(t)}\,dt+(\pi/2)^{1/2}\left( \int_{0}^{\pi /2}|f(t)-g(t)|\,dt\right) ^{1/2}\\ & \leq\int_{0}^{\pi/2}\sqrt{g(t)}\,dt+(\pi/2)^{1/2}\varepsilon^{1/2}% \end{align*} and similarly, $$ \int_{0}^{\pi/2}\sqrt{g(t)}\,dt\leq\int_{0}^{\pi/2}\sqrt{f(t)}\,dt+(\pi /2)^{1/2}\varepsilon^{1/2}% $$ and so \begin{align*} \left( \frac{1}{\pi/2}\int_{0}^{\pi/2}\sqrt{f(t)}\,dt\right) ^{2} & \leq\left( \frac{1}{\pi/2}\int_{0}^{\pi/2}\sqrt{g(t)}\,dt+\frac {\varepsilon^{1/2}}{(\pi/2)^{1/2}}\right) ^{2}\\ & \leq\left( \frac{1}{\pi/2}\int_{0}^{\pi/2}\sqrt{g(t)}\,dt\right) ^{2}+\frac{\varepsilon}{(\pi/2)}+\frac{2\varepsilon^{1/2}}{\pi/2}\int_{0}% ^{\pi/2}\sqrt{g(t)}\,dt\\ & \leq\left( \frac{1}{\pi/2}\int_{0}^{\pi/2}\sqrt{g(t)}\,dt\right) ^{2}+\frac{\varepsilon}{(\pi/2)}+\frac{2\varepsilon^{1/2}}{\pi/2}\left( \int_{0}^{\pi/2}\sqrt{f(t)}\,dt+(\pi/2)^{1/2}\right) \\ & \leq\left( \frac{1}{\pi/2}\int_{0}^{\pi/2}\sqrt{g(t)}\,dt\right) ^{2}+\frac{\epsilon}{3}% \end{align*} provided $\varepsilon$ is taken small enough. Now combine these two inequalities.