Uses Hölder’s inequality to turn strong
convergence of matrix fields into convergence of their determinant
integrals.
Statement
Let
,
let
be a measure space, and let
be any filter on an index set
.
Let
,
equipped with the operator norm induced by
on
.
Suppose
,
eventually
along
,
and
Then
along the same filter.
Assumptions
The dimension is positive. Membership in
includes almost-everywhere strong measurability and finite seminorm. The
measure need not be finite, and the filter need not be a sequence
filter. No pointwise convergence or common pointwise bound is
assumed.
Conclusion
The limiting determinant is integrable, the approximating
determinants are eventually integrable, and their integrals converge. In
fact their differences converge to zero in
.
The supplied convergence predicate records both seminorm convergence
and eventual membership; the separate hypothesis on
is also necessary for the proof. The one-dimensional case is handled
directly, without using the undefined exponent
at
.
Proof route
Apply the determinant difference estimate, then Hölder with exponents
and
.
The resulting bound contains the factor
,
which tends to zero; the other factor stays bounded by the triangle
inequality. Treat
directly.
Proof steps
Start with the pointwise estimate and
integrability. The determinant bound gives, at every
,
For any field
,
comparison with the zero operator gives
Continuity of determinant supplies measurability, so
is integrable. Apply this to
and, eventually, to
.
Hence the determinant difference is eventually integrable. All following
estimates are on this eventual set of indices.
If
,
no nontrivial Hölder exponent is needed. Step 1 reduces to
If
,
write Hölder’s inequality with the actual functions. Its
conjugate exponents are
and
.
The functions
and
belong to
by the hypotheses and Minkowski’s inequality. Thus
The second inequality is
Hölder. The next is Minkowski applied to the two scalar norm functions.
The last uses
.
This identifies every function and exponent in the estimate without
introducing new names for their norms.
Take the limit and then subtract the integrals.
Since
,
eventually it is at most
.
The last bound in Step 3 is then at most
The coefficient is finite because
.
Together with Step 2 this proves
convergence for every
.
The integrability checked in Step 1 now justifies
All eventual statements
and limits use the same filter
;
neither finite total measure nor a sequence index was assumed.