
Relaxed Controls 75
Integrating both sides of this inequality from t
0
to t and using (3.6.24) gives
log ρ(t) ≤ log
α +
Z
t
t
0
µρds
≤ log α +
Z
t
t
0
µds.
From this we get (3.6.25).
If α = 0, then (3.6.24) holds for all α
1
> 0. Hence (3.6.25) holds for all
α
1
> 0. Letting α
1
→ 0 now yields ρ(t) ≡ 0. Hence (3.6.25) is trivially true.
Remark 3.6.7. The proof shows that if α > 0 and strict inequality holds in
(3.6.24), then strict inequality holds in (3.6.25).
Theorem 3.6.8. Let I be a compact interval in R
1
, let X be a compact
interval in R
n
, and let R = I×X. Let G = R×U, where U is a region of R
m
,
and let f be a continuous mapping from G to R
n
. Let Ω be a mapping