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By Juan Carlos Cabello Píñar

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26 Capítulo 1. Espacios normados. 3. Relación de ejercicios 1. Pruébese que si X es de dimensión finita, entonces todo funcional alcanza su norma. 2. Pruébese que todo subconjunto compacto de un espacio normado de dimensión infinita tiene interior vacío. 3. Sean X un espacio normado, M un subespacio vectorial de X. Pruébese que si M es de dimensión finita entonces, para cada x ∈ X, existe m ∈ M tal que x − m = dist(x, M ). 4. Sea X un espacio normado de dimensión finita y M un subespacio cerrado no total de X.

Para cada p ∈ R con p ≥ 1, definimos: 1 Lp [0, 1] = {f ∈ [0, 1] → K : f es medible Lebesgue , |f |p < ∞}. 0 Dados f, g ∈ Lp [0, 1] y t ∈ [0, 1] se tiene: |f (t) + g(t)|p ≤ 2p m´ax{|f (t)|p , |g(t)|p } ≤ 2p (|f (t)|p + |g(t)|p ), 1 por tanto 0 |f + g|p < ∞ y f + g ∈ Lp [0, 1]. De forma similar, αf ∈ Lp [0, 1] para todo f ∈ Lp [0, 1] y α ∈ K. Por tanto, Lp [0, 1] es un subespacio vectorial de K[0,1] . 5. Subespacios complementados. Cociente de espacios normados. 31 (que se obtiene de forma similar a su anteriormente comentada versión numérica) permite deducir que la función νp : Lp [0, 1] → R definida por 1 νp (f ) = |f | p 1 p (f ∈ Lp [0, 1]) 0 es una seminorma.

Supongamos que T (BX ) es un entorno de cero en Y. Entonces T es abierta. 3 (Teorema de la aplicación abierta de Banach) Toda aplicación lineal, continua y sobreyectiva entre dos espacios de Banach es abierta. El Teorema de la aplicación abierta admite la siguiente reformulación equivalente.

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Análisis Funcional by Juan Carlos Cabello Píñar


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