Let a>0, A={f∈C^2 ([0,a],R) , f(0)=f′(0)=0} N_1 (f)= sup{∣f(x)∣+∣f′′(y)∣ ,x,y∈[0,a]} N_2 (f)=sup{∣f(x)+f′′(x)∣ ,x∈[0,a]} Prove that N_1 and N


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Question Number 124415 by snipers237 last updated on 03/Dec/20
$Let a>0, A={f∈C^2 ([0,a],R) , f(0)=f′(0)=0} N_1 (f)= sup{∣f(x)∣+∣f′′(y)∣ ,x,y∈[0,a]} N_2 (f)=sup{∣f(x)+f′′(x)∣ ,x∈[0,a]} Prove that N_1 and N_2 are equivalents norms$
$L e t a > 0, A = {f \in C^{2} ([0, a], R), f (0) = f^{'} (0) = 0}$ $N_{1} (f) = s u p {∣ f (x) ∣ + ∣ f^{″} (y) ∣, x, y \in [0, a]}$ $N_{2} (f) = s u p {∣ f (x) + f^{″} (x) ∣, x \in [0, a]}$ $P r o v e t h a t N_{1} a n d N_{2} a r e e q u i v a l e n t s n o r m s$
Commented bymnjuly1970 last updated on 03/Dec/20

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