# Abel summability of integrals

In Introduction to Fourier Analysis on Euclidean Spaces by Stein and Weiss, there is the following footnote on page 5 (the integral referred to is the integral of $\sin(x)/x$) on $[0,\infty)$):

As is well known, in this case the limit $\lim_{p\rightarrow\infty} \int_0^p f(x)\ dx$ exists. It is an easy exercise to show that whenever $f$ is locally integrable and such a limit, $l$, exists the Abel means $A_\epsilon =\int_0^\infty e^{-\epsilon x}f(x) \ dx$ converge to $l$.

The limit is of course taken as $\epsilon$ tends to $0$, and we are using Lesbegue integrals.

Unfortunately, I am not finding this to be an easy exercise. Break each interval $[n,n+1]$ into $2^n$ pieces. Consider the function that is alternately $2^{n/2}$ and $-2^{n/2}$ on these intervals. This is conditionally convergent, but the Abel means don't converge for small $\epsilon$.

Either I made a mistake with this counterexample, or the theorem requires more hypotheses. Could I please get a few hints?

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And the integral is at most $1/(2^{n/2})$ away from the limit, so it converges as $n$ goes to infinity. – Potato Aug 13 '12 at 7:23
You could expand on the reason why you think the Abel means do not converge. – Did Aug 13 '12 at 7:55
@did A function is integrable only if it is absolutely integrable. So we are looking at the Abel means of a function which is $2^{n/2}$ on the interval $[n,n+1]$ - essentially the exponential $\sqrt 2^{x}$. Now if we set $\epsilon$ equal to say, 1/100 in $A_\epsilon$, it obviously does not converge. – Potato Aug 13 '12 at 7:59
This probably means that S&W define $A_\epsilon$ as the limit when $p\to\infty$ of the integral on $(0,p)$ of you-know-which-function. – Did Aug 13 '12 at 8:04
@did Perhaps. Do you have a reference on this? It would certainly be a more interesting theorem if we got absolute integrability with some additional mild hypothesis. – Potato Aug 13 '12 at 8:06

No further assumption is required if we understand the defining integral in Laplace transform as an improper one. Indeed, the following proposition holds.

Proposition. Assume $f$ is locally integrable on $(0, \infty)$ and the improper integral

$$\int_{0}^{\infty} f(x) \; dx := \lim_{p\to\infty} \int_{0}^{p} f(x) \; dx$$

exists. Then its Laplace transform

$$A_{s} = \mathcal{L}f(s) := \int_{0}^{\infty} f(x)\,e^{-sx} \; dx = \lim_{p\to\infty} \int_{0}^{p} f(x)\,e^{-sx} \; dx$$

exists for $s \geq 0$ and

$$\lim_{s\downarrow 0}A_{s} = A_{0} = \int_{0}^{\infty} f(x) \; dx.$$

For the proof, let $h(x) = \int_{0}^{x} f(t) \; dt$. Then $h(x)$ is absolutely integrable, and is bounded since the limit $h(\infty) := \lim_{p\to\infty} h(p)$ exists. By integration by parts, we have

\begin{align*} \int_{0}^{p} f(x)\,e^{-sx}\;dx &= \left[h(x)\,e^{-sx}\right]_{0}^{p} + s \int_{0}^{p} h(x) \, e^{-sx} \; dx \\ &= h(p)\,e^{-sp} + s \int_{0}^{p} h(x) \, e^{-sx} \; dx. \end{align*}

Here, note that

$$|h(x) \, e^{-sx}| \leq \| h \|_{L^{\infty}} e^{-sx}.$$

Thus by Lebesgue's dominated convergence theorem, we have

$$\lim_{p\to\infty} \int_{0}^{p} f(x)\,e^{-sx}\;dx = s \int_{0}^{\infty} h(x) \, e^{-sx} \; dx.$$

This proves the existence of $A_{s}$. To prove that $A_{s} \to A_{0}$ as $s \downarrow 0$, we observe that

$$F(0) = h(\infty) = s \int_{0}^{\infty} h(\infty) \, e^{-sx} \; dx.$$

Thus we have

\begin{align*} |A_{s} - A_{0}| &\leq s \int_{0}^{\infty} |h(x) - h(\infty)| \, e^{-sx} \; dx \\ &= \int_{0}^{\infty} |h(u/s) - h(\infty)| \, e^{-u} \; du. \qquad(u = sx) \end{align*}

Since this integrand is bounded by the dominating function $2\|h\|_{L^{\infty}} \,e^{-u}$, we can apply Lebesgue's dominated convergence theorem again, yielding

$$\lim_{s\downarrow 0} |A_{s} - A_{0}| \leq \int_{0}^{\infty} \lim_{s\downarrow 0} |h(u/s) - h(\infty)| \, e^{-u} \; du = 0.$$

Therefore $A_{s} \to A_{0}$ as desired.

Of course, in the reference book the Laplace transform is defined as a Lebesgue integral sense. Even in this case, only a mild auxiliary condition, such as

$$f(x) = O(e^{\epsilon x}) \quad \text{for all} \ \epsilon > 0$$

or

$$\int_{0}^{x} |f(t)| \; dt = O(e^{\epsilon x}) \quad \text{for all} \ \epsilon > 0$$

guarantees the proposition, since we only need to show that the improper integral defining $A_{s}$ in fact converges in a Lebesgue sense. Actually, as you read that page, you will find that the book also requires some conditions on $f$ so that $A_{\epsilon}$ is defined for all $\epsilon > 0$.

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I think the proof given by sos440 requires $f$ to be continuous, since we need to take the derivative of $h$ in the integation by parts. To generalize the proof to the case that $f$ is only locally integrable, one can prove first $h$ is differentiable in the $L^1_{loc}$ sense, and the derivative is just $f$. Then we generalize also the integration by parts to show $$\int_0^\infty f(x)e^{-\epsilon x} dx = \epsilon \int_0^\infty h(x) e^{-\epsilon x} dx$$ The rest of the proof remains the same.

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