If
\Phi:[0,\infty )\rightarrow \mathbb{R}
Φ
:
[
0
,
∞
)
→
R
is convex and continuous with
\Phi (0)=0
Φ
(
0
)
=
0
and if
q\in (1,\infty )
q
∈
(
1
,
∞
)
,
q^{\prime }:=\frac{q}{q-1}
q
′
:
=
q
q
−
1
, we first prove that the inequality
\Phi \left( \int_{0}^{\infty }f(r)dr\right) \leq C\int_{0}^{\infty }f(r)\Phi ^{\prime }(r^{1/q^{\prime }})dr
Φ
(
∫
0
∞
f
(
r
)
d
r
)
≤
C
∫
0
∞
f
(
r
)
Φ
′
(
r
1
/
q
′
)
d
r
for every
f\in L^{q}(0,\infty)
f
∈
L
q
(
0
,
∞
)
,
f\geq 0
f
≥
0
with
||f||_{q}\leq 1
∣
∣
f
∣
∣
q
≤
1
holds when
C=1
C
=
1
. In general, both sides may be
\pm \infty
±
∞
. Related inequalities for
f\in L^{1}(\mathbb{R}^{N})\cap L^{q}(\mathbb{R}^{N})
f
∈
L
1
(
R
N
)
∩
L
q
(
R
N
)
,
f\neq 0
f
≠
0
are derived. This inequality is independent of Jensen's inequality and, when
q=\infty
q
=
∞
, it is an elaboration on an inequality of Steffensen which was discussed elsewhere by the author. The next goal of the paper is to identify the range of the admissible constants
C
C
and, in particular, to characterize the optimal constant when
\Phi \geq 0
Φ
≥
0
or
\Phi \leq 0
Φ
≤
0
. It turns out that
C=1
C
=
1
is “almost always” optimal, at least in a restricted sense, but not always when
q<\infty
q
<
∞
: Given
q
q
, the admissible constants lie on an interval containing
1
1
whose left (right) endpoint is the supremum (infimum) of a function defined on some (left/right dependent) subset of
\mathbb{R}^{2}
R
2
. If
q=2
q
=
2
, these extrema can be calculated in a number of examples. Among other things, this reveals that
C=1
C
=
1
need not be optimal when
\Phi \geq 0
Φ
≥
0
and
\Phi _{+}^{\prime }(0)=0
Φ
+
′
(
0
)
=
0
or when
\Phi \leq 0
Φ
≤
0
and
\Phi _{+}^{\prime} (0)=-\infty
Φ
+
′
(
0
)
=
−
∞
.