Download Analysis II: Funktionen mehrerer Variablen by Friedmar Schulz PDF

By Friedmar Schulz

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Example text

AN ∈ K mit N K ⊂ ⋃ Uak . k=1 Sei δ (ε) ∶= 12 min { δ (ε, a1 ), . . , δ (ε, aN ) } > 0. Seien x, x′ ∈ K mit ∣x − x′ ∣ < δ. Dann gibt es ein k ∈ { 1, . . , N } mit x ∈ Uak , das heißt ∣x − ak ∣ < 12 δ (ε, ak ). 5 Stetige Funktionen und Abbildungen auf kompakten Mengen weshalb also 53 ε ε ∣f (x′ ) − f (ak )∣ < , ∣f (x) − f (ak )∣ < , 2 2 ∣f (x) − f (x′ )∣ < ε für alle x, x′ ∈ K, ∣x − x′ ∣ < δ. Folglich ist f gleichmäßig stetig auf K. 4 Satz. Sei K ⊂ Rn kompakt und f ∶ K → Rm eine stetige Abbildung.

Hnlich gilt für alle , k ∈ N0 , > k, dass ∣x − xk ∣ ≤ ∣x − x −1 ∣ + ⋯ + ∣xk+2 − xk+1 ∣ + ∣xk+1 − xk ∣ −1 ≤ ∑ Lm ∣x1 − x0 ∣ . m=k Weil die geometrische Reihe konvergiert, ist die Folge (xk )k∈N0 also eine CauchyFolge. Wegen der Abgeschlossenheit von F gilt xk → x ∈ F . Für → ∞ folgt, dass ∞ Lk ∣x − xk ∣ ≤ ∑ Lm ∣x1 − x0 ∣ = ∣x1 − x0 ∣ 1−L m=k wie behauptet. 4 Bemerkung. Der Kontraktionssatz gilt auch in einem vollständig normierten Vektorraum, das heißt in einem Banachraum und heißt dann Banachscher Fixpunktsatz.

1 Funktionen und Abbildungen 31 für x = (x1 , . . , xn ) ∈ Rn . αn xα1 1 ⋅ . . ⋅ xαnn . Im Fall k = 2 gilt zum Beispiel n n i,j =1 i=1 P (x) = ∑ aij xi xj + ∑ bi xi + c. (vi) σk ∶ Rn → R, σk (x) ∶= ∑ 1≤i1 <⋯

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