Let $b>1$. Find the sup of $\int_0^1 g(t^b)dt$, where the sup runs over all continuous function $g$ such that $\int_0^1 |g|^b \leq 1$.
My attempt: $\int_0^1 g(t^b)dt=\int_0^1 g(s)\frac{1}{b}s^{\frac{1}{b}-1}ds$. One can use Holder inequality to dominate it. But $\int_0^1 s^{-1}ds=\infty$, so we just are led to the integral $\leq \infty$. Nonsense!
For $0 < c < 1$ consider the function $$ g(t) = \left( \frac{1-c}{t^c}\right)^{1/b} \, . $$ Then $$ \int_0^1 g(t)^b \, dt = \int_0^1 \frac{1-c}{t^c} \, dt = 1 $$ and $$ \int_0^1 g(t^b) = \int_0^1 \frac{(1-c)^{1/b}}{t^c} \, dt = (1-c)^{1/b-1} \, . $$ Since $1/b-1 < 0$, the right-hand side becomes arbitrarily large for $c \to 1$, i.e. the supremum is $\infty$.