Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN: 9780470458365
Author: Erwin Kreyszig
Publisher: Wiley, John & Sons, Incorporated
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Request explain the marked portion of closed graph theorem proof

10.2 Closed graph theorem (Banach, 1932) Let X and Y be Banach spaces and F: X→ Y be a closed linear map. Then F is continuous. Proof: By condition (i) of 6.2, it is enough to prove that F is bounded on some neighborhood of 0. For each positive integer n, let V₁ = {x € X: ||F(x)|| ≤ n}. We prove that some V₁ contains a neighborhood of 0 in X. Now X = UV₂ = UVA, n=1 n=1 where V, denotes the closure of the set V₁ in X. Hence (V₂) = 0, n=1 where (V₁) denotes the complement of the set V₁ in X. Since X is a Banach space, one of the open sets (V₂) must not be dense in X by Baire's theorem (3.4). Hence we find a positive integer p, some ro € X and 8 >0 such that U(ro, 6) CVp. We shall show that U(0,6) C Vap- First, note that U(0,8) CV2p. For, if z X with ||*|| < 8, then x+xo € U(xo, 8) CVp. Also, 2o € Vp. If (vn) and (w) are sequences in V, such that Un →+To and wn →To, then Un Un-Wn € V₂p since →x, where Request Explain ||F(vn - wn)| ≤|F(vn)|| + ||F(wn)|| ≤ 2p. Thus x € V2p. In particular, for every z EU (0,8), there is some 2₁ in V2p such that ||*-*₁|| < 8/2. Hence U(0, 6) C V2p + U(0,6).
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Transcribed Image Text:10.2 Closed graph theorem (Banach, 1932) Let X and Y be Banach spaces and F: X→ Y be a closed linear map. Then F is continuous. Proof: By condition (i) of 6.2, it is enough to prove that F is bounded on some neighborhood of 0. For each positive integer n, let V₁ = {x € X: ||F(x)|| ≤ n}. We prove that some V₁ contains a neighborhood of 0 in X. Now X = UV₂ = UVA, n=1 n=1 where V, denotes the closure of the set V₁ in X. Hence (V₂) = 0, n=1 where (V₁) denotes the complement of the set V₁ in X. Since X is a Banach space, one of the open sets (V₂) must not be dense in X by Baire's theorem (3.4). Hence we find a positive integer p, some ro € X and 8 >0 such that U(ro, 6) CVp. We shall show that U(0,6) C Vap- First, note that U(0,8) CV2p. For, if z X with ||*|| < 8, then x+xo € U(xo, 8) CVp. Also, 2o € Vp. If (vn) and (w) are sequences in V, such that Un →+To and wn →To, then Un Un-Wn € V₂p since →x, where Request Explain ||F(vn - wn)| ≤|F(vn)|| + ||F(wn)|| ≤ 2p. Thus x € V2p. In particular, for every z EU (0,8), there is some 2₁ in V2p such that ||*-*₁|| < 8/2. Hence U(0, 6) C V2p + U(0,6).
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