Consider $[0,1]$ with its Lebesgue measure. Let $1<p<\infty$ and set $$\Gamma = \left\{f\in L^p([0,1]): \int_0^1 5f(x)x^3dx \leq \frac{1}{\pi}\int_0^1 f(x) dx.\right\}$$ Prove that $\Gamma$ is closed in $L^p([0,1])$.
I ran into this problem while working through old qual problems and am pretty stuck. I have been trying to show that $\Gamma^C$ is open, i.e. for all $f \in \Gamma^C$, there exists $\delta>0$ such that $B_\delta(f) \subset \Gamma^C$. If we let $f \in \Gamma^C$, then we need to find some $\delta > 0$ such that if $\|f-g\|_p = \left(\int_0^1 |f-g|^p dx\right)^\frac{1}{p} < \delta$, then $g \in \Gamma^C$. My thought was to split up the integral: $$ \frac{1}{\pi} \int_0^1 g dx = \frac{1}{\pi} \int_0^1 (g -f) dx + \frac{1}{\pi} \int_0^1 f dx < \frac{1}{\pi} \int_0^1 (g -f) dx + \int_0^1 5f(x)x^3 dx.$$ However, I'm not sure where to go from here. Does anyone have any suggestions?
Hint: Suppose $f_1,f_2,\dots \in \Gamma,$ and $f_n\to f $ in $L^p.$ Then $f_n\to f$ in $L^1.$ Doesn't that imply the given inequality holds for $f?$