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Hummingbird25
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Hi
I'm tasked with proving the following:
Let S be an open interval S and f: S -> \mathbb{R}^n be a continuous function.
Let || \cdot || be norm on \mathbb{R}^n. show
1) There exist a K > 0 such that ||x|| \leq K||x||_1 ; \ x \in \mathbb{R}^n, ||x||_1 = \sum_{i=1} ^n |x_i|.
My Solution:
According to the definition the norm of a vector x in R^n is the non-negative scalar ||x|| = \sqrt{x_1^{2} + x_2^{2} + \cdots x_n^2}
The L1-norm can be written as ||x||_1 = |x_1| + |x_2| + \cdots + |x_n|
Expanding the inequality:
\sqrt{x_1^{2} + x_2^{2} + \cdots x_n^2} \leq K|x_1| + K|x_2| + \cdots + K|x_n|
Is it then concludable from this that since K>0, then the right side of the inequality sign will always be larger than the left side?
Sincerley Yours
Humminbird25
I'm tasked with proving the following:
Let S be an open interval S and f: S -> \mathbb{R}^n be a continuous function.
Let || \cdot || be norm on \mathbb{R}^n. show
1) There exist a K > 0 such that ||x|| \leq K||x||_1 ; \ x \in \mathbb{R}^n, ||x||_1 = \sum_{i=1} ^n |x_i|.
My Solution:
According to the definition the norm of a vector x in R^n is the non-negative scalar ||x|| = \sqrt{x_1^{2} + x_2^{2} + \cdots x_n^2}
The L1-norm can be written as ||x||_1 = |x_1| + |x_2| + \cdots + |x_n|
Expanding the inequality:
\sqrt{x_1^{2} + x_2^{2} + \cdots x_n^2} \leq K|x_1| + K|x_2| + \cdots + K|x_n|
Is it then concludable from this that since K>0, then the right side of the inequality sign will always be larger than the left side?
Sincerley Yours
Humminbird25
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