Find a function that has finite values for lower norm but becomes infinite I recently came across the following question: does there exist a non-decreasing function $h : [0,1) \rightarrow \Re^+$, i.e. with non-negative range, that satisfies $\|h\|_1=1$ and $\|h\|_a\leq 1$, where $a$ is some value in $[1,2)$, i.e. for lower orders of norm, but has $\|h\|_2 = \infty$? 
This seems quite tricky, and I had scratched my head for quite a while on this. Wonder if anyone might have an idea if this is trivial or difficult. thanks!

Such a function cannot exist.
Assume that there exists a measurable function $f : [0,1) \to \mathbb{R}^+$ such that $\|f\|_a \le 1$, $\forall a \in [1,2)$.
Notice that $f^a \le f^b$ for any $a \le b$ with $a, b \in [1,2)$. Pick an increasing sequence $(a_n)_{n=1}^\infty$ in $[1,2)$ which converges to $2$. Then $(f^{a_n})_{n=1}^\infty$ is an increasing sequence of nonnegative functions so we can use the Lebesgue Monotone Convergence Theorem:
$$1 \ge \lim_{n\to\infty}\int_0^1 f(x)^{a_n}\,dx= \int_0^1 \big(\lim_{n\to\infty} f(x)^{a_n}\big)\,dx = \int_0^1 f(x)^2\,dx$$
Hence, it also must be $\|f\|_2 \le 1$.
