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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!

mechanodroid
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Phi
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1 Answers1

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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 $\|f\|_2 \le 1$.

mechanodroid
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  • Oh...! I withdraw my statement that $||f||_a \leq 1$ cannot be satisfied. This is surprising to me....! – Phi Nov 17 '17 at 22:37
  • @Phi I made a mistake, it's not $\le 1$ even with this modified example. – mechanodroid Nov 17 '17 at 22:41
  • @Phi Have a look now. It seems that $p \mapsto |f|_p$ is a continuous function, so it cannot suddenly jump to $+\infty$. See here and here. – mechanodroid Nov 17 '17 at 23:09
  • @mechanodriod This seems to suggest that in order to have $||f||_2=\infty$, any such function must satisfy $||f||_a \leq c$ with $c$ increases in $a$. The previous function that you constructed is like that. Do you think it is possible to identify the tightest c for each $a$ such that $||f||_2 = \infty$? – Phi Nov 18 '17 at 01:19
  • I just officially made this as a question "Find the lowest value for a lower norm subject to norm-2 being infinite". Any thought along this line is very much appreciated! appears to be much harder though – Phi Nov 18 '17 at 01:47