推 hau :想了非常久的難題，謝謝 01/01 10:41
※ 引述《hau》之銘言：
: assume μ(X) = 1
: Prove lim ║f║_p = exp [∫ log｜f｜dμ ].
: p→0 X
: positive part of log｜f｜is integrable
: if ∫ log｜f｜dμ= -∞
: X
: in this case lim║f║_p = 0
: p→0
[我看不懂你的提示，下面有個方法應該可以解決之。]
Assume μ(X) ＝ 1, and that ∥f∥_r ＜ ∞ for some r ＞ 0. Prove that
lim ∥f∥_p ＝ exp {∫ log |f| dμ} if exp {-∞} is defined to be 0.
p→∞ X
Proof.
Notice that lim ∥f∥_p ＝ exp {lim (1/p)‧log (∫_X |f|^p dμ)}, we pay
attention to lim (1/p)‧log (∫_X |f|^p dμ):
(1) Since -log x is convex, we know that
-log (∫_X |f|^p dμ) ≦ ∫_X -log |f|^p dμ by Jensen's inequality.
It implies that
∫_X log |f| dμ ≦ (1/p)‧log (∫_X |f|^p dμ).
So, ∫_X log |f| dμ ≦ lim inf (1/p)‧log (∫_X |f|^p dμ).
(2) Consider
(1/p)‧log (∫_X |f|^p dμ) ≦ (1/p)(∫_X |f|^p dμ - 1) by 1＋x ≦ e^x,
|f|^p - 1
＝ ∫ ------------- dμ since μ(X) ＝ 1,
X p
:＝ ∫_{|f|≧1} ＋ ∫_{|f|＜1}.
^^^^^^^^^^^^ ^^^^^^^^^^
(A) (B)
For (A): It is not hard to see that (|f|^p - 1)/p ↓ log |f| as p↓0, and
f in L^r(X) with μ(X) ＝ 1, the dominated function can be (|f|^r - 1)/r,
by the inequality (|f|^r - 1)/r ≧ (|f|^p - 1)/p (≧0), where p≦r. So,
with help of LDCT (or MCT), we have
lim ∫_{|f|≧1} (|f|^p - 1)/p dμ ＝ ∫_{|f|≧1} log |f| dμ.
p→0
For (B): It is not hard to see (|f|^p - 1)/p ↑log |f| (≦0) as p↓0.
That is, "-"(|f|^p - 1)/p ↓ "-" log |f|. From - (|f|^r - 1)/r in L,
say p≦r, we get, by LDCT (or MCT),
lim ∫_{|f|＜1} - (|f|^p - 1)/p dμ ＝ ∫_{|f|＜1} -log |f| dμ,
p→0
i.e., ∫_{|f|＜1} (|f|^p - 1)/p dμ → ∫_{|f|＜1} log |f| dμ.
We conclude that
∫_X log |f| dμ ≧ lim sup (1/p)‧log (∫_X |f|^p dμ)
from two parts (A), and (B).
Finally, we complete it via (1) and (2). □
NOTE. 出處: Real And Complex Analysis-Water Rudin, exer. 5-(d), p. 71.
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※ 編輯: math1209 來自: 220.133.4.14 (12/28 15:22)