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Eftersom ε(h) = g(a + h) − g(a) − Dg(a)(h) og b = g(a), f˚as at
Df(b)(g(a + h) − g(a)) + δ(k) =
Df(g(a))(Dg(a)(h) + ε(h)) + δ(k)
Udtrykket Df(g(a)) ganges ind i parentesen:
Df(g(a))(Dg(a)(h) + ε(h)) + δ(k) =
Df(g(a))(Dg(a)(h)) + Df(g(a))(ε(h)) + δ(k)
Da T = Df g(a) Dg(a) og b = g(a), f˚as at
Det følger heraf, at
Df(g(a))(Dg(a)(h) + Df(g(a))(ε(h)) + δ(k) =
T (h) + Df(b)(ε(h)) + δ(k)
f(g(a + h)) − f(g(a)) − T (h) = Df(b)(ε(h)) + δ(k)
For at gøre beviset færdigt sættes
Nu skal det blot vises, at
T1(h) = Df(b)(ε(h)) T2(h) = δ(k)
Tj(h)
||h||
Da Df(b) er en lineær operator, gælder at
Det vides, at ε(h)
||h||
T1(h)
||h||
→ 0, n˚ar h → 0 for j = 1, 2.
= Df(b)(ε(h))
||h||
= Df(b)
ε(h)
||h||
→ 0 n˚ar h → 0. Enhver lineær operator afbilleder 0 over i
0, jf. proposition 3.1 i [Axl97]. Eftersom der ifølge [Che] gælder, at Df(b) er
kontinuert i 0, f˚as at
ε(h)
Df(b) → 0, n˚ar h → 0
||h||
For at bevise at T2(h)
||h||
Derfor bestemmes normen af k:
||k||
→ 0, n˚ar h → 0, skal det først vises, at ||h|| er begrænset.
||k|| = ||g(a + h) − g(a)||
Da ε(h) = g(a + h) − g(a) − Dg(a)(h), f˚as at
||k|| = ||Dg(a)(h) + ε(h)||
Trekantsuligheden anvendes p˚a dette udtryk:
||Dg(a)(h) + ε(h)|| ≤ ||Dg(a)(h)|| + ||ε(h)||