Sea (V, || · ||) un espacio vectorial normado. Prueba |||x||...

Transcript

00:01 Be a normed vector space, we prove that the absolute value of the norm of x minus the norm of y is less or equal to the norm of x minus y for all vectors x and y in the normed vector space b, that is, we take two elements whatever in the normed vector space b and we will prove this this inequality.

00:36 Let's start with the norm of x and the norm of x is equal to the norm of x minus y plus y simply subtracting and adding y within the norm and now we apply the triangular inequality of the norm and this is less or equal to the norm of the vector x minus y plus the norm of the vector y.

01:09 This is the triangular inequality.

01:14 Very well then from here subtracting on both sides the inequality of the norm of y we have that the norm of x minus the norm of y is less or equal than the norm of x minus y let's say that this inequality is the inequality 1 and on the other hand we can do the following operation the norm of of y is equal to the norm of y minus x plus x again what i did was add and subtract x within the norm therefore it is exactly the same now this because it is triangular again this is less or equal to the norm of the vector y minus x plus the norm of the vector x then triangular inequality now it means that the norm of y is less or equal to the norm of y minus x plus the norm of x now this is the same as the following this is not my y less or equal to the norm of the scalar minus 1 by x minus y norm of that vector plus the norm of x where we simply use the scale minus 1 and show that it is the same to do the operation y minus x that minus 1 by x minus y if we distribute this minus 1 within the parentheses we get just this therefore we have an expression this norm that is exactly equal to this other now i apply the property of the norm that tells me that the the norm of a scalar by a vector, which is what we have here, is equal to the absolute value of the scalar, the absolute value of minus 1, by the norm of the vector.

03:40 Therefore, the inequality is identical to this other...

Question

Sea (V, || · ||) un espacio vectorial normado. Prueba |||x|| - ||y||| ? ||x - y||

Question

Sea (V, || · ||) un espacio vectorial normado. Prueba |||x|| - ||y||| ? ||x - y||

Please give Ace some feedback