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I am looking for a measurable function $f\colon (0;1) \to [0;\infty)$ with $$ \int_0^1 f(x) \mathrm dx < \infty, \quad \int_0^1 f(x)^{1+\alpha} \mathrm dx = \infty \quad \text{for all}\, \alpha > 0. $$ Taking $f(x) = x^{-1}$ does not work because the first statement is violated.

Taking $f(x) = x^{-1+\beta}$ for $\beta > 0$ small does not work because the second statement is violated for small $\alpha$.

Taking $f(x) = x^{-1+x}$ does not work because the first statement is violated (according to Wolfram Alpha).

Or can we prove that such a function does not exist? It seems like $p \mapsto \Vert f\Vert_p$ is continuous in the sense that the pre-image of $\{\infty\}$ is closed.

Kolodez
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    Might be that I'm confusing something, but doesn't $f(x) = \frac{1}{x\ln^2(x)}$ on $(0,\frac{1}{2})$ and $0$ otherwise, do the job? – Dominik Kutek Jul 11 '21 at 10:24
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    Obviously there are some $f_n$ supported on $[2^{-n},2^{-(n-1)}]$ which is $L^1$ but not $L^{1+1/n}$. Scale appropriately and patch everything together. – user10354138 Jul 11 '21 at 10:27
  • @DominikKutek Sorry, yes you are absolutely right. It is very much possible for a function to be integrable for $0 < p \leq 1$ and not $p>1$, and your function qualifies. I got my explanation of that wrong, indeed the endpoint $1$ must be contained in the set of $p$ for which the norm is finite. – Sarvesh Ravichandran Iyer Jul 11 '21 at 10:28
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    I think @DominikKutek's example works. For $\alpha \geqslant 0$, the substitution $x = e^t$ gives $$\int_0^\infty (f(x))^{1 + \alpha} \ {\mathrm d}x = \int_{-\infty}^{-\log 2} e^{-\alpha t} \frac{1}{(t^2)^{1 + \alpha}} \ {\mathrm d}t.$$ The above will be finite iff $\alpha = 0$. – Aryaman Maithani Jul 11 '21 at 11:29
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    The answer of user zhw gives you more than what you asked, but it is good to have it for future reference $f \in L^1$, but $f \not\in L^p$ for all $p > 1$ – Mittens Jul 11 '21 at 15:08

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