Mwanafunzi, weka swali linalokusumbua ili walimu na wazoefu wakusaidie

Mwanafunzi, weka swali linalokusumbua ili walimu na wazoefu wakusaidie

Hello. Wanajukaa natumai wote ni wazimaa wa afya, mimi ni mwanafunzi nasomea BD in wildlife management, kunaswali hilo tumepewa nahitaji ufafanuzi namna yakulifanya
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Hello I hope your doing well I was asking for anyone who can help me giving me some information about petroleum chemistry including working fields and other projects
 
gammaparticles said:
Kati ya Yai na Kuku,
Kipi kilianza kuwepo duniani?
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Kuku alianza..kuku ndio kiumbe na yai ili liwe kuku lazima kuku alitamie iyo kisayansi

Kidini pia kama MUNGU alimuumba Adam hakuumba mimba basi na pia aliumba kuku hakuumba yai.

Kitu katika full developed form kinaweza kabili mazingira tofauti na MUNGU labda angeumba mtoto mchanga angeishi vipi dunuani au yai lingetotoa vipi bila mama.
 
Wakuu nnaomba ujuzi wa namna ya kuweka ramani ya study area..Napata shida namna ya ku produce ramani ya Kata ya Dakawa, Mvomero kwa vijiji vya Wami Sokoine, Wami Luhindo, Wami Dakawa, Milama na Kwamuhunzi. Msaada tafadhali
 
Sawa. Kaswali kangu haka hapa Student Room - Page 12:

Let (X) be a compact metric space and (V) be a strongly continuous semigroup of contractions on (L^p(X)) for some (1 \leq p < \infty). Assume that (V) is irreducible, i.e., there is no closed subspace of (L^p(X)) invariant under all operators in (V).

Prove that if there exists a function (f \in L^p(X)) such that (V(t)f \to 0) weakly in (L^p(X)) as (t \to \infty), then the spectrum of the generator (A) of (V(t)) is not discrete.
 
TensorFlow said:
Let (X) be a compact metric space and (V) be a strongly continuous semigroup of contractions on (L^p(X)) for some (1 \leq p < \infty). Assume that (V) is irreducible, i.e., there is no closed subspace of (L^p(X)) invariant under all operators in (V).

Prove that if there exists a function (f \in L^p(X)) such that (V(t)f \to 0) weakly in (L^p(X)) as (t \to \infty), then the spectrum of the generator (A) of (V(t)) is not discrete.
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Kwa msaada wa.........


To prove that the spectrum of the generator AAA of the strongly continuous semigroup V(t)V(t)V(t) is not discrete under the given conditions, we start with the definition of the generator and some properties of the semigroup.

Definitions and Setup​

  1. Generator of the Semigroup: The generator AAA of the semigroup V(t)V(t)V(t) is defined by
    Af=lim⁡t→0+V(t)f−ftAf = \lim_{t \to 0^+} \frac{V(t)f - f}{t}Af=t→0+limtV(t)f−f
    for fff in the domain D(A)D(A)D(A) of AAA.
  2. Weak Convergence: The assumption that V(t)f→0V(t)f \to 0V(t)f→0 weakly in Lp(X)L^p(X)Lp(X) means that for any g∈Lp′(X)g \in L^{p'}(X)g∈Lp′(X) (where 1p+1p′=1\frac{1}{p} + \frac{1}{p'} = 1p1+p′1=1),
    ⟨V(t)f,g⟩→0as t→∞.\langle V(t)f, g \rangle \to 0 \quad \text{as } t \to \infty.⟨V(t)f,g⟩→0as t→∞.
  3. Irreducibility: The semigroup V(t)V(t)V(t) being irreducible means there are no closed subspaces of Lp(X)L^p(X)Lp(X) that are invariant under V(t)V(t)V(t) for all t≥0t \geq 0t≥0.

Proof​

Now, we will show that if there exists f∈Lp(X)f \in L^p(X)f∈Lp(X) such that V(t)f→0V(t)f \to 0V(t)f→0 weakly in Lp(X)L^p(X)Lp(X), then the spectrum of the generator AAA is not discrete.
  1. Spectrum and Weak Convergence: Since V(t)f→0V(t)f \to 0V(t)f→0 weakly, this implies that the semigroup V(t)V(t)V(t) is "decaying" in the direction of the function fff. The weak convergence to zero suggests that 000 is in the spectrum of the generator.
  2. Spectrum of the Generator: The spectrum σ(A)\sigma(A)σ(A) can be viewed as the set of λ∈C\lambda \in \mathbb{C}λ∈C such that A−λIA - \lambda IA−λI is not invertible. In our context, for the spectrum to be discrete, the resolvent (A−λI)−1(A - \lambda I)^{-1}(A−λI)−1 would need to exist for λ\lambdaλ in some neighborhood of every point in the spectrum.
  3. Non-Discrete Spectrum: The weak convergence V(t)f→0V(t)f \to 0V(t)f→0 suggests that the semigroup has a "non-trivial" behavior at infinity. More importantly, due to irreducibility, if 000 were an isolated point of the spectrum, we would be able to find an invariant subspace corresponding to 000, contradicting the irreducibility assumption. Thus, 000 cannot be isolated.
  4. Conclusion: Because 000 is not isolated in the spectrum of AAA, it follows that the spectrum cannot be discrete. Therefore, we conclude that the spectrum of the generator AAA is not discrete.
In summary, the presence of a function f∈Lp(X)f \in L^p(X)f∈Lp(X) for which V(t)f→0V(t)f \to 0V(t)f→0 weakly implies that the generator's spectrum has non-isolated points, leading to the conclusion that it is not discrete. This completes the proof.
 
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