Absztrakt: | We study cosine and sine Fourier transforms defined by $F(t):=
(2/\pi) \int^\infty_0 f(x) \cos tx dx$ and $\widetilde F(t):=
(2\pi) \int^\infty_0 f(x) \sin tx dx$, where $f$ is $L^1$-integrable
over $[0, \infty)$. We also assume that $F$ and $\widetilde F$
are locally absolutely continuous over $[0, \infty)$. In particular,
this is the case if both $f(x)$ and $xf(x)$ are $L^1$-integrable
over $[0, \infty)$. Motivated by the inversion formulas, we consider
the partial integrals $s_\nu(f, x):= \int^\nu_0 F(t)\cos xt dt$
and $\widetilde s_\nu(f, x):= \int^\nu_0 \widetilde F(t)\sin
xt dt$ and the modified partial integrals $u_\nu(f, x):= s_\nu(f,
x)- F(\nu)(\sin \nu x)/x$ and $\widetilde u_\nu(f, x):= \widetilde
s_\nu(f, x)+ \widetilde F(\nu)(\cos \nu x)/x$, where $\nu> 0$.
We give necessary and sufficient conditions for the $L^1[0, \infty)$-convergence
of $u_\nu(f)$ and $\widetilde u_\nu(f)$ as well as for the $L^1[0,
X]$-convergence of $s_\nu(f)$ and $\widetilde s_\nu(f)$ to $f$
as $\nu\to \infty$, where $0< X< \infty$ is fixed. On the other
hand, in certain cases we conclude that $s_\nu(f)$ and $\widetilde
s_\nu(f)$ cannot belong to $L^1[0, \infty)$. Consequently, it
makes no sense to speak of their $L^1[0, \infty)$-convergence
as $\nu\to \infty$.\par
As an intermediate tool, we use the
Ces\`aro means of Fourier transforms. Then we prove Tauberian
type results and apply Sidon type inequalities in order to obtain
Tauberian conditions of Hardy-Karamata kind.\par
We extend these
results to the complex Fourier transform defined by $$G(t):=
{1 \over (2\pi)} \int^\infty_{-\infty} g(t) e^{-ixt} dt,$$ where
$g$ is $L^1$-integrable over $(- \infty, \infty)$. |