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Let $f: [0, 1] \to \mathbb R$ be an $L^1$ function. Define for each $r > 0$, the blow up $f_r:[0, 1] \to \mathbb R$ by

$$f_r (x) := \frac{f(rx)}{r}.$$

Suppose $f_r$ converges in $L^1$ to some function $g$ as $r \to 0^+$. Is it true that $g$ agrees a.e. with a linear function?

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Yes. In this situation, $F_r(x)=\int_0^x f_r(t)\, dt \to G(x)=\int_0^x g(t)\, dt$ pointwise, and thus \begin{align*} G(x) & = \lim_{r\to 0+} \frac{1}{r} \int_0^{x} f(rt)\, dt = \lim_{r\to 0+} \frac{x}{r} \int_0^1 f(rxs)\, ds \\ &=x^2\lim_{p\to 0+} \frac{1}{p}\int_0^1 f(ps)\, ds = x^2G(1). \end{align*}

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  • $\begingroup$ Very nice idea to consider the anti derivative! $\endgroup$
    – Nate River
    Commented Dec 22, 2023 at 19:34
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    $\begingroup$ @NateRiver: Thank you, but come to think of it, it's actually unnecessary since also $f_r\to g$ pointwise a.e. on a subsequence and similarly $g(x)=\lim f(rx)/r = (x/x_0) \lim f(px_0)/p = cx$. $\endgroup$
    – Christian Remling
    Commented Dec 22, 2023 at 19:44
  • $\begingroup$ Ah yes, you can just take subsequences twice, and they both converge to $g$. As an aside, one can probably prove similarly in $\mathbb R^n$ that the only possible blow up limits are planes and cones. $\endgroup$
    – Nate River
    Commented Dec 22, 2023 at 20:28

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