complex-analysislebesgue-integralreal-analysis
I am quite confused about the notion of integrability. In the context of an introduction to complex analysis and Fourier transforms, I am told that if $f$, a complex or real valued function, satisfies the following:
$$ \int_{-\infty}^{\infty} \lvert f(x) \rvert dx<\infty$$
then it is absolutely integrable. However, does this also imply that $f$ is integrable? What can we conclude about $f$ given the above condition on its absolute value (or modulus). I have no notions of Lebesgue integrability.
In finite dimensions, all norms on a vector space are equivalent. Thus we tend to stick with the well-understood Euclidean norm on $\Bbb R^n$, occasionally pulling out the sup or taxicab norms when computationally convenient.
But most function spaces are infinite-dimensional, which means norms finally have enough freedom to disagree with each other. And the thing is, there are several different norms that are useful. And when as here, we are talking about several possible spaces, each with its own norm, it becomes confusing which norm is being referred to.
There are only two notations for norms that are so widely used that an author can use them without explaining. These are:
In specialized fields of mathematics, there are some other notations that are well-known in that field, and may be used when speaking to those well-versed in the field. But when speaking to a wider mathematically inclined audience, the notations above are the only ones that can safely be used without explanation. And of course even then, there will always be a few novices who might need an introduction.
The tip-off that Daniel Fischer is correct about the intent here is the definition specifies that $\sum_{n=1}^\infty \|f_n\|$ must converge uniformly. If $\|f_n\|$ meant the supremum norm of $f_n$, then the series would be a sum of real numbers, and talking about "uniform" convergence makes no sense. Uniform with respect to what? In order to talk about uniform convergence, $\sum \|f_n\|$ must be a sum of functions. The interpretation that makes $\|f_n\|$ a function is that it means the function $$\|f_n\| : S \to \Bbb R : x \mapsto \|f_n(x)\|$$
This is a really nice question, I was surprised after thinking for a couple of hours and none of the standard examples in a harmonic analysis course fitting the description. In any case, I will try to present what can and what cannot be done to the integrability of the Fourier transform of a Lipschitz function.
Notice also that I will be employing standard analysis notation (such as $\lesssim$ and $O(g)$) as well as some standard theorems in Fourier analysis, such as van der Corput's lemma for bounding oscillatory integrals (also known as 'stationary phase method'). If you are not familiar with those, I would gladly refer you to Stein's "Harmonic Analysis: Real-Variable Methods, Orthogonality, and Oscillatory Integrals", whose chapter VIII contains all that we shall use here.
1.1. If instead of $f$ being Lipschitz, one requires the different condition that $\|f(x+h)-f(x)\|_{L^2(dx)} = O(h^{\alpha}),$ for some $\alpha > 1/2,$ then the same conclusion of Bernstein's theorem holds. That is, if the above condition is satisfied and $f \in L^1(\mathbb{R}),$ then $\widehat{f} \in L^1(\mathbb{R}).$ A proof of this fact can be found, for instance, in the book "Introduction to the Theory of Fourier Integrals", by Titchmarsh.
As a by-product, notice that if $f \in L^1(\mathbb{R})$ is Lipschitz and has compact support, then the condition in Titchmarsh's theorem above is trivially satisfied from the Lipschitz condition, and thus $\widehat{f} \in L^1(\mathbb{R}).$ This shows that, if a counterexample exists, it cannot have compact support.
1.2. On the other hand, notice that $f \in L^1, f' \in L^{\infty}$ implies directly that $f \in L^{\infty},$ and thus, by interpolation, $f \in L^p, \forall p \ge 1.$ By the Hausdorff--Young inequality, we have that $\widehat{f} \in L^q, \forall q \ge 2.$ In fact, this can be significantly improved:
Claim: Let $f \in L^1(\mathbb{R})$ be Lipschitz, i.e., $f' \in L^{\infty}.$ Then $\widehat{f} \in L^p(\mathbb{R}), \forall p > 1.$
Proof: In Grafakos's "Modern Fourier Analysis", the following result can be found as a by-product of a characterization of Lipschitz spaces: 'Let $\Delta_j(f) = (\psi(\xi/2^j) \widehat{f})^{\vee},$ where $\psi$ is a smooth bump supported on $[1/2,4]$ and equal to 1 on $[1,2].$ If $f$ is Lipschitz, then
$$\sup_{j \ge 0} 2^j \|\Delta_j(f)\|_{\infty} < + \infty.'$$
We will not prove this here, but by assuming such a result, we have the following: for $\psi$ as in the statement of the Littlewood-Paley characterization of Lipschitz functions, we have
$$\int_{2^j}^{2^{j+1}} |\widehat{f}(\xi)|^p \, d\xi \le \int_{\mathbb{R}} |\psi(\xi/2^j) \widehat{f}(\xi)|^p \psi(\xi/2^j) \, d \xi \lesssim 2^{(1-\frac{p}{2})j} \|\Delta_j(f)\|_2^{p},$$
where in the last inequality we used Hölder and Plancherel. Now, by interpolation we have
$$\|\Delta_j(f)\|_2 \le \|\Delta_j(f)\|_{\infty}^{1/2} \|\Delta_j(f)\|_1^{1/2}.$$
From the definition of $\Delta_j,$ we have $\|\Delta_j(f)\|_1 \lesssim \|f\|_1,$ and from the characterization of Lipschitz functions mentioned before, $\|\Delta_j(f)\|_{\infty} \lesssim 2^{-j}.$ Putting all those considerations together, we obtain
$$ \int_{2^j}^{2^{j+1}} |\widehat{f}(\xi)|^p \, d\xi \lesssim 2^{(1-p)j},$$
and the right-hand side above is summable in $j \ge 1$ if $p>1,$ as desired. $\square.$
Thus, the Fourier transform of any such counterexample has to integrate to any order $p>1.$
$$ \widehat{f}(x) = \frac{e^{i\frac{x^2}{2\log x}}}{x \log x} \psi_0(x).$$
Then we need to prove that
$$f(\xi) = \int_{\mathbb{R}} \frac{e^{i\frac{x^2}{2\log x} - i\xi x}}{x \log x} \psi_0(x) \, dx$$
is Lipschitz and integrable.
2.1. $f$ is Lipschitz: it is not too hard (even though it is a bit technical) to show that $f$ is Lipschitz if and only if the integrals
$$ J_N(\xi) = \int_{10}^N \frac{e^{i\frac{x^2}{2 \log x}}}{\log x} \psi_0(x) e^{-i \xi x} \, dx $$
are uniformly bounded on $\xi, N.$ Define then the phase function $\phi(x,\xi) = \frac{x^2}{2\log x} - \xi x.$ Suppose that, for some $x_{\xi} \in \mathbb{R},$ we have $\partial_x \phi(x_{\xi},\xi) = 0.$ As $\partial_x \phi(x,\xi) = \frac{x}{\log x} + \frac{x}{2 \log^2 x} - \xi,$ then it is also not hard to show that $x_{\xi} \gtrsim |\xi| \log |\xi|.$ We split then the integral defining $J_N(\xi)$ into the one over the interval $[5,N] \setminus [\xi^{1/2},\xi^{3/2}],$ denoted by $J_N^1(\xi),$ and the one over $[\xi^{1/2},\xi^{3/2}],$ denoted by $J_N^2(\xi).$ Thus, $J_N(\xi) = J_N^1(\xi) + J_N^2(\xi).$
For $J_N^1,$ we use that the derivative of the phase satisfies $|\partial_x\phi(x,\xi)| \gtrsim \max\{|\xi|,\frac{x}{\log x}\},$ while for $x$ sufficiently large (10 should do for this) $\partial_{xx}^2 \phi(x,\xi) > 0,$ so we are in position to apply van der Corput's lemma, which then implies that
$$|J_N^1(\xi)| \lesssim \frac{1}{|\xi|}.$$
For the analysis of $J_N^2,$ we notice that the interval $[\xi^{1/2},\xi^{3/2}]$ contains the zero $x_{\xi}$ of $\partial_x \phi(x,\xi),$ and in that interval it also holds that $\partial_{xx}^2 \phi(x,\xi) \gtrsim \frac{1}{\log \xi}.$ Using van der Corput's lemma again, we have that
$$|J_N^2(\xi)| \lesssim \frac{(\log \xi)^{1/2}}{\log \xi} \lesssim \frac{1}{(\log \xi)^{1/2}}.$$
This finishes the case $\xi \gg 1.$ For the small $\xi$ case, the analysis here carries out even easier than what is written here. Thus, $f$ is Lipschitz.
2.2. $f$ is integrable: The idea is basically the same as here, but integrating by parts first. Indeed, we need to prove some uniform estimates on the functions
$$f_N(\xi) := \int_{10}^N \frac{e^{i(\frac{x^2}{2\log x} - \xi x)}}{x \log x} \psi_0(x) \, dx.$$
Integrating by parts with factors $u = \frac{e^{i\frac{x^2}{2\log x}}}{x \log x} \psi_0(x)$ and $dv = e^{-i \xi x} dx.$ This gives that $f_N(\xi)$ is equivalently equal to
$$\frac{1}{i\xi}\frac{ e^{i( \frac{N^2}{2\log N} - \xi N)}}{N \log N} + $$ $$ \frac{1}{i \xi} \int_{10}^N \partial_x \left(\frac{x^2}{2\log x}\right) x \log x \frac{e^{i(\frac{x^2}{2\log x} - \xi x)} }{x^2 \log^2 x} \, \psi_0(x) dx + $$ $$ \frac{1}{ i\xi} \int_{10}^N \frac{\log x + 1}{x^2 \log^2 x} e^{i(\frac{x^2}{2 \log x} - \xi x)} \psi_0(x) \, dx + $$ $$ \frac{1}{i\xi} \int_{10}^N \frac{e^{i(\frac{x^2}{2\log x} - \xi x)}}{x \log x} \psi_0'(x) \, dx =: \frac{1}{i\xi} (I_1(N,\xi) + I_2(N,\xi) + I_3(N,\xi) + I_4(N,\xi)).$$
One notices quite immediately that $I_1(N,\xi) \to 0$ as $N \to \infty,$ so we do not need to worry about that term when taking limits. Also, $\frac{\psi_0'(x)}{x \log x}$ has support on $[10,20]$ and is smooth there, while the derivative of the phase $\partial_x\phi(x,\xi)$ does not vanish for $\xi$ small, and neither does it on the support for $\xi$ sufficiently large. Thus, this term is bounded and decays as fast as one wishes - i.e., $|I_4(N,\xi)| = O_R(\xi^{-R})$ when $\xi \to \infty,$ for any $R > 0.$ Thus $I_4(N,\cdot) \in L^1(\mathbb{R})$ uniformly on $N \to \infty.$
We then analyze the remaining terms $I_2,I_3.$ In fact, the analysis of $I_2$ is much more delicate than that of $I_2,$ so we perform only the former and defer the latter to whoever wishes to fill in the gaps.
Rewrite $I_2(N,\xi)$ as
$$\int_{10}^N \left( \frac{x}{\log x} + \frac{x}{2 \log^2x} \right) \frac{x \log x}{x^2 \log^2 x} e^{i(\frac{x^2}{2\log x} - \xi x)} \psi_0(x) \, dx = $$ $$ = 2 \int_{10}^N \frac{e^{i(\frac{x^2}{2\log x} - \xi x)}}{ (\log x)^2} \psi_0(x) dx + \int_{10}^N \frac{e^{i(\frac{x^2}{2\log x} - \xi x)}}{(\log x)^3} \psi_0(x) \, dx.$$
These resemble a lot the integrals we had to estimate for proving $f$ is Lipschitz. In fact, the very same argument is applicable in these cases as well, where one obtains (after redoing the same computations with van der Corput and what not)
$$|I_2(N,\xi)| \lesssim \frac{1}{(\log |\xi|)^{3/2}}.$$
As mentioned before, the analysis of $I_3$ is in fact easier, and doing the computations yields
$$|I_3(N,\xi)| \lesssim \frac{1}{|\xi|}.$$
Notice that all such estimates were made uniformly on $N \to \infty.$ Putting all together and taking limits, one obtains that, for $|\xi| \gg 1,$
$$|f(\xi)| \lesssim \frac{1}{|\xi| (\log|\xi|)^{3/2}}.$$
As the right-hand side is integrable and $f$ was already seen to be Lipschitz, this implies $f \in L^1(\mathbb{R}),$ as desired.
Best Answer
A function $f$ is Lebesgue integrable iff its integral exists and $|\int f| <\infty$. However, an equivalent definition is the one you have give above.
For $f$ with a defined (Lebesgue) integral by definition we have $\int f = \int f^+ - \int f^-$. If $f$ is in addition integrable it then $\int f^+ <\infty$ and $\int f^- < \infty$ so $\int f^+ + \int f^- = \int |f| < \infty$ so where I have used the fact that $|f| =f^+ + f^ - $. Thus $|f|$ is integrable as all non-negative functions have a defined integral.
Conversely $|f|$ is integrable then $f^+ \leq |f|$ and $f^- \leq |f|$ so $ \int f^+ < \int |f| < \infty $ and $ \int f^- < \int |f| < \infty $ which implies both the negative and positive parts of $f$ are finite and the integral is thus just the difference of two positive real numbers, so is finite. Thus $f$ is integrable.