where the infimum is taken over all u satisfying 42.7.36and 42.7.37.
Note that a norm on A_{0}× A_{1} would be
||(a,a )|| ≡ max(||a || ,t||a || )
0 1 0 A0 1 A1
and so J
(t,⋅)
is the restriction of this norm to the subspace of A_{0}× A_{1} defined by
{(a,a) : a ∈ A0 ∩ A1}
. Also for each t > 0 J
(t,⋅)
is a norm on A_{0}∩A_{1} and furthermore,
any two of these norms are equivalent. In fact, for 0 < t < s,
J (t,a) = max (||a|| ,t||a|| )
( A0 A1 )
≥ max ||a||A0 ,s||a||A1
= J (s,a)
( s )
≥ max t ||a||A0 ,s||a||A1
= smax (||a|| ,t||a|| )
t A0 A1
≥ sJ (t,a) .
t
The following lemma is significant and follows immediately from the above
definition.
Lemma 42.7.2Suppose a ∈
(A ,A )
0 1
_{θ,q,J}and a = ∫_{0}^{∞}u
(t)
dt
t
where u is describedabove. Then letting r > 1,
{ ( )
u (t) ≡ u(t) if t ∈ 1r,r .
r 0 otherwise
it follows that
∫ ∞
ur(t) dt∈ A0 ∩ A1.
0 t
Proof: The integral equals ∫_{1∕r}^{r}u
(t)
dt
t
.∫_{1∕r}^{r}
1
t
dt = 2lnr < ∞. Now u_{r } is
measurable in A_{0}∩ A_{1} and bounded. Therefore, there exists a sequence of measurable
simple functions,
{sn}
having values in A_{0}∩ A_{1} which converges pointwise and
uniformly to u_{r}. It can also be assumed J
(r,sn(t))
≤ J
(r,ur(t))
for all t ∈
[1∕r,r]
.
Therefore,
∫
lim r J (r,s − s ) dt = 0.
n,m→ ∞ 1∕r m n t
It follows from the definition of the Bochner integral that
∫ ∫
r dt r dt
lnim→∞ 1∕rsnt = 1∕rur t ∈ A0 ∩ A1.
This proves the lemma.
The remarkable thing is that the two spaces,
(A0,A1)
_{θ,q} and
(A0,A1)
_{θ,q,J} coincide
and have equivalent norms. The following important lemma, called the fundamental
lemma of interpolation theory in [?] is used to prove this. This lemma is really
incredible.
Lemma 42.7.3Suppose for a ∈ A_{0} + A_{1}, lim_{t→0+}K
(t,a)
= 0 and lim_{t→∞}
K(t,a)-
t
= 0.Then for any ε > 0, there is a representation,
is nondecreasing and so if its limit is
positive, the integrand would have a non integrable singularity like t^{−θq−1}. Next consider
what happens to
K(t,a)
t
as t →∞.
Claim:t →
K-(t,a)
t
is decreasing.
Proof of the claim:Choose a_{0}∈ A_{0} and a_{1}∈ A_{1} such that a_{0} + a_{1} = a
and
K (t,a) + εt > ||a || + t||a ||
0 A0 1 A1
let s > t. Then
K-(t,a)+-tε ≥ ||a0||A0-+-t||a1||A1≥ ||a0||A0 +-s||a1||A1-≥ K-(s,a).
t t s s
Since ε is arbitrary, this proves the claim.
Let r ≡ lim_{t→∞}
K(t,ta)
. Is r = 0? Suppose to the contrary that r > 0. Then the
integrand of 42.7.43, is at least as large as
q−1 K (t,a) q−1
t−θqK (t,a) ---t---≥ t−θqK (t,a) r
−θq q−1 q q(1−θ)− 1
≥ t (tr) r ≥ r t
whose integral is infinite. Therefore, r = 0.
Lemma 42.7.3, implies there exist u_{i}∈ A_{0}∩ A_{1} such that a = ∑_{i=−∞}^{∞}u_{i},
the convergence taking place in A_{0} + A_{1}with the inequality of that Lemma
holding,