Let's say $x(\cdot)$ is a continuous function over $t$. The problem is to find a $x$ such that
$\int_{-1}^1 t^3 x(t) dt$ is maximized. The constraint is $\int_{-1}^1|x(t)|^3 dt \leq 2$. I was trying to use Euler-Lagrange equation to find an $x$. The problem is how to develop a Euler-Lagrange equation for this problem. Would Holder's inequality be of any help?
We can use Hölder's inequality here rather than Euler Lagrange.
Take $L^3[-1,1]$ and define $f(x) = \int_{-1}^1 t^3 x(t) dt $.
We see that $f$ is linear, and the constraint is $\|x\|_3 \le \sqrt[3]{2}$.
The problem becomes: $\max \{ f(x) | \|x\|_3 \le \sqrt[3]{2} \}$.
Hölder's inequality gives $|f(x)| \le \|\phi\|_{3 \over 2} \| x\|_3$, where $\phi(t) = t^3$.
Note that $\|\phi\|_{3 \over 2} = \sqrt[3]{4^2 \over 11^2}$. We have equality iff $ | {x(t) \over \|x\|_3 }|^3 = |{ \phi(t) \over \| \phi \|_{3 \over 2} } |^{3 \over 2}$ for ae. $t \in [-1,1]$.
Letting $x(t) = (\operatorname{sgn} t)\sqrt[3]{22 \over 4}\sqrt{|t^3|}$, we obtain $\max \{ f(x) | \|x\|_3= \sqrt[3]{2} \}= \sqrt[3]{2^5 \over 11^2}$.
Since $x$ is continuous, it solves the original problem.