Lp space has been described in detail in functional analysis (but I have not learned it in detail), so here is a little more repetition.
1. Lp space definition
Let ( X , F , μ ) (X,\mathscr F,\mu)(X,F,μ ) is a metric space, defining a function on which the power of absolute value p can be integrated (p ≥ 1 p\ge1p≥1 ) The entire set isL p ( X , F , μ ) L^p(X,\mathscr F,\mu)Lp(X,F,μ ) . That is,L p L^pLThe function in p
satisfies: ∫ X ∣ f ∣ pd μ < ∞ \int_X |f|^pd\mu < \infty∫X∣f∣pdμ<∞
2. Lp space is a Banach space
To demonstrate this conclusion, first demonstrate that Lp is a normed space, and then prove it completely.
To demonstrate that it is a normed space, non-negativity, positive definiteness and linearity are obviously, mainly to demonstrate the triangle inequality. Prove the triangle inequality, that is, prove the Minkowski inequality , that is:
for 1 ≤ p < ∞ , ∀ f , g ∈ L p 1\le p < \infty,\forall f,g \in L^p1≤p<∞,∀f,g∈Lp ,equation:
∣ ∣ f + g ∣ ∣ p ≤ ∣ ∣ f ∣ ∣ p + ∣ ∣ g ∣ ∣ p ||f+g||_p\le||f||_p+||g||_p∣∣f+g∣∣p≤∣∣f∣∣p+∣∣g∣∣p
To prove Minkowski's inequality, you must first prove Holder's inequality :
For a pair of conjugate numbers ( 1 < p , q < ∞ : 1 q + 1 p = 1 1<p,q<\infty:\frac{1}{q }+\frac{1}{p}=11<p,q<∞:q1+p1=1), ∀ f , g ∈ L p \forall f,g \in L^p ∀f,g∈Lp ,equation:
∣ ∣ fg ∣ ∣ ≤ ∣ ∣ f ∣ ∣ p + ∣ ∣ g ∣ ∣ q ||fg||\le||f||_p+||g||_q∣∣fg∣∣≤∣∣f∣∣p+∣∣g∣∣q
And Holder's inequality depends on the following lemma :
for a pair of conjugate numbers p , qp,qp,q,对 ∀ a , b ≥ 0 \forall a,b \ge 0 ∀a,b≥0,有:
a 1 / p b 1 / q ≤ a p + b q a^{1/p}b^{1/q}\le \frac{a}{p}+\frac{b}{q} a1/pb1/q≤pa+qb
The logic chain is like this, the proof is omitted here, and it can be found everywhere.
Two special p-values are discussed below: p = ∞ p=\inftyp=∞和 0 < p < 1 0<p<1 0<p<1.
p = ∞ p=\infty p=∞ , define the norm:
∣ ∣ f ∣ ∣ ∞ = inf { a ∈ R + , μ ( ∣ f ∣ > a ) = 0 } ||f||_{\infty}=\inf\{a\ in \R^+,\mu(|f|>a)=0\}∣∣f∣∣∞=inf{
a∈R+,m ( ∣ f ∣>a)=0 } , it can be proved that it is indeed a norm.
0 < p < 1 0 < p < 10<p<When 1 , define:
∣ ∣ f ∣ ∣ p = ∫ X ∣ f ∣ pd μ ||f||_p=\int_X|f|^pd\mu∣∣f∣∣p=∫X∣f∣p dμ, note that unlike the traditional definition of the p-norm, it does not take the p-th root anymore. So obviously, it no longer satisfies the linear property,so when0 < p < 1 0<p<10<p<At 1 , it is no longer a normative space.
It can be proved that 0 ≤ p ≤ ∞ 0\le p\le \infty0≤p≤∞ ,L p L^pLThe Cauchy column in p- space converges, so: 0 < p < 1 0<p<10<p<1 o'clockL p L^pLp space is a complete distance space,1 ≤ p ≤ ∞ 1\le p\le \infty1≤p≤∞ is a Banach space.
3. Average convergence
Before explaining the definition of average convergence, first prove the following theorem. I think the proof of this theorem is relatively comprehensive, so it is included as follows.
Theorem Let 0 < p ≤ ∞ 0<p\le\infty0<p≤∞ , if{ fn } ⊂ L p \{f_n\}\subset L^p{
fn}⊂Lp satisfies:
lim n , m → ∞ ∣ ∣ fn − fm ∣ ∣ p = 0 \lim_{n,m\to\infty}||f_n-f_m||_p=0n,m→∞lim∣∣fn−fm∣∣p=0
, there existsf ∈ L pf\in L^pf∈Lp , such that:
lim n → ∞ ∣ ∣ f − fn ∣ ∣ p = 0 \lim_{n\to\infty}||f-f_n||_p=0n→∞lim∣∣f−fn∣∣p=0
is now called{ fn } \{f_n\}{
fn} (order p) averagely converges tofff , written as:fn → L pf f_n \stackrel{L_p}{\rightarrow}ffn→Lpf
Proof:
\space
Theorem: Relationship between average convergence and rest convergence
Average convergence is actually a relatively strong condition.
(1) If fn → L pf f_n \stackrel{L_p}{\rightarrow}ffn→Lpf ,则fn → μ f f_n \stackrel{\mu}{\rightarrow}ffn→mf且 ∣ ∣ f n ∣ ∣ p → ∣ ∣ f ∣ ∣ p ||f_n||_p\to||f||_p ∣∣fn∣∣p→∣∣f∣∣p;
(2)若 f n → μ f f_n \stackrel{\mu}{\rightarrow}f fn→mf或 f n → a . e . f f_n \stackrel{a.e.}{\rightarrow}f fn→to . and .f,则 f n → L p f f_n \stackrel{L_p}{\rightarrow}f fn→Lpf。
The proof
of (2) is more technical, and the current level is limited, only (1) is proved.