In mathematical analysis , the Young's inequality for integral operators , is a bound on the
L
p
→
L
q
{\displaystyle L^{p}\to L^{q}}
operator norm of an integral operator in terms of
L
r
{\displaystyle L^{r}}
norms of the kernel itself.
Statement
Assume that
X
{\displaystyle X}
and
Y
{\displaystyle Y}
are measurable spaces,
K
:
X
×
Y
→
R
{\displaystyle K:X\times Y\to \mathbb {R} }
is measurable and
q
,
p
,
r
≥
1
{\displaystyle q,p,r\geq 1}
are such that
1
q
=
1
p
+
1
r
−
1
{\displaystyle {\frac {1}{q}}={\frac {1}{p}}+{\frac {1}{r}}-1}
. If
∫
Y
|
K
(
x
,
y
)
|
r
d
y
≤
C
r
{\displaystyle \int _{Y}|K(x,y)|^{r}\,\mathrm {d} y\leq C^{r}}
for all
x
∈
X
{\displaystyle x\in X}
and
∫
X
|
K
(
x
,
y
)
|
r
d
x
≤
C
r
{\displaystyle \int _{X}|K(x,y)|^{r}\,\mathrm {d} x\leq C^{r}}
for all
y
∈
Y
{\displaystyle y\in Y}
then
∫
X
|
∫
Y
K
(
x
,
y
)
f
(
y
)
d
y
|
q
d
x
≤
C
q
(
∫
Y
|
f
(
y
)
|
p
d
y
)
q
p
.
{\displaystyle \int _{X}\left|\int _{Y}K(x,y)f(y)\,\mathrm {d} y\right|^{q}\,\mathrm {d} x\leq C^{q}\left(\int _{Y}|f(y)|^{p}\,\mathrm {d} y\right)^{\frac {q}{p}}.}
Particular cases
Convolution kernel
If
X
=
Y
=
R
d
{\displaystyle X=Y=\mathbb {R} ^{d}}
and
K
(
x
,
y
)
=
h
(
x
−
y
)
{\displaystyle K(x,y)=h(x-y)}
, then the inequality becomes Young's convolution inequality .
See also
Young's inequality for products
Notes
Theorem 0.3.1 in: C. D. Sogge , Fourier integral in classical analysis , Cambridge University Press, 1993. ISBN 0-521-43464-5
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