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Let $\newcommand{\RR}{\mathbb{R}}f: \RR \to \RR$ be a twice differentiable function that is in $\mathcal{L}^1$. Suppose that also $f'' \in \mathcal{L}^1$. Can we conclude that $f' \in \mathcal{L}^1$?

What about the case where we know $f \in \mathcal{L}^1$ and $f^{(m)} \in \mathcal{L}^1$, can we say something about the integrability of $f^{(k)}$ for $k \leq m$ ?

In general we can't say that the derivative of a smooth integrable function is integrable, but maybe the integrability of $f''$ helps?

K.Power
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Carlos Esparza
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  • The intuition here definitely seems right. I assume someone will chime in here with some sobolev inequality that gives an affirmative answer, at least to the first question. – Jake Mirra Oct 15 '20 at 18:16
  • You might try to connect the integrability of the $f'(x)$ to the decay of its Fourier transform and see that it has the desired decay because the Fourier transforms of $f$ and $f''(x)$ must. – user293794 Oct 15 '20 at 18:25
  • @user293794 That's an interesting idea. I tried to use it but got stuck because I don't see how to apply the Fourier inversion theorem (we don't know that $\hat{f}$ or $\hat{f''}$ are in $L^1$...) – Carlos Esparza Oct 15 '20 at 18:48
  • Right you would want some sort of lemma like "if $f$ is $L^1$ and differentiable and $\xi \hat{f}(\xi)$ is $L^\infty$ then $f'$ is $L^1$. I believe there should be a more direct way of seeing this based on FTC, but I'm not sure yet. – user293794 Oct 15 '20 at 19:34
  • Somewhat related: https://math.stackexchange.com/q/3766807/168433 – md2perpe Oct 16 '20 at 10:39
  • @MartinR Yes this was precisely what I was looking for :) I just didn't have enough time to read you answer in detail. – Carlos Esparza Oct 30 '20 at 04:54

1 Answers1

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If $f''$ is continuous (or more generally, if $f'$ is absolutely continuous) then $$ f(x+1) = f(x) + f'(x) + \int_x^{x+1} (x+1-t)f''(t) \, dt $$ from Taylor's theorem with the integral remainder. It follows that $f' \in {\cal L}^1$ if $$ g(x) = \int_x^{x+1} (x+1-t)f''(t) \, dt $$ is in ${\cal L}^1$ and that can be shown with the Fubini-Tonelli theorem: $$ \int_{-\infty}^{\infty} |g(x)| \, dx \le \int_{-\infty}^{\infty}\left(\int_x^{x+1} (x+1-t)|f''(t)| \, dt \right) dx \\ = \int_{-\infty}^{\infty} \left( \int_{t-1}^t (x+1-t)|f''(t)| \, dx \right) dt \\ = \int_{-\infty}^{\infty} |f''(t)|\left( \int_{t-1}^t (x+1-t) \, dx \right) dt \\ = \frac 12 \int_{-\infty}^{\infty} |f''(t)| \, dt < \infty \, . $$

This answers the first part of your question.

The same approach works with higher derivatives: If $f$ and $f^{(n+1)}$ are in ${\cal L}^1$ and $f^{(n+1)}$ is continuous (or more generally, $f^{(n)}$ is absolutely continuous), then we can use Taylor's theorem for $n$ distinct non-zero offsets $0 < h_1 < h_2 < \ldots < h_n$: $$ f(x+h_j) = f(x) + \frac{f'(x)}{1!} h_j + \frac{f''(x)}{2!} h_j^2 + \ldots + \frac{f^{(n)}(x)}{n!} h_j^n + \int_x^{x + h_j} \frac{f^{(n+1)}(t)}{n!} (x+h_j - t)^n \, dt $$ for $j=1, \ldots, n$. For a given $k$ with $1 \le k \le n$ one can find coefficients $c_1, \ldots, c_n$ such that $$ c_1 h_1^l + c_2 h_2^l + \ldots + c_n h_n^l = \begin{cases} 1 & \text{ for } l=k \, , \\ 0 & \text{ for } l=1, \ldots, n, l \ne k \, . \end{cases} $$ That works because the determinant of the coefficient matrix of that linear equation system is $h_1 h_2 \cdots h_n$ multiplied with the Vandermonde determinant of $h_1, h_2, \ldots, h_n$ and therefore non-zero.

It follows that $$ \frac{f^{(k)}(x)}{k!} = \sum_{j=1}^n c_j \left( f(x+h_j) - f(x) - \int_x^{x + h_j} \frac{f^{(n+1)}(t)}{n!} (x+h_j - t)^n \, dt \right) $$ and the same reasoning as above shows that $f^{(k)} \in \cal L^1$.

Carlos Esparza
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Martin R
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