Question #126922

If a sub additive functional defined on a normed space X is nonnegative outside a sphere {x Illxll = r}, show that it is nonnegative for all x E X


1
Expert's answer
2020-07-21T18:26:24-0400

Suppose that ff is a sub additive functional on XX. Then, by definition, we have

x,yXf(x+y)f(x)+f(y).\forall x,y \in X \quad f(x+y)\leq f(x)+f(y). For any xXx\in X outside a sphere f(x)0f(x)\geq0 . Suppose that x~\tilde{x} lies either inside a sphere or on its boundary. This means that x~=αr||\tilde{x}||=\alpha\leq r. We consider two cases:

  1. α>0\alpha>0 . We set n=[rα]+2n=[\frac{r}{\alpha}]+2 , where [][] denotes the truncation. Then nα>rn\alpha>r . It means, that nx~>r||n\tilde{x}||>r and nx~n\tilde{x} lies outside the sphere. Using sub additivity, we get 0f(nx~)=f((n1)x~+x~)f(x~)+f((n2)x~+x~)...nf(x~)0\leq f(n\tilde{x})=f((n-1)\tilde{x}+\tilde{x})\leq f(\tilde{x})+f((n-2)\tilde{x}+\tilde{x})\leq...\leq n\,f(\tilde{x})

Therefore, we have f(x~)0f(\tilde{x})\geq0 .

2. α=0\alpha=0 . Using the definition of the norm, we have x~=0\tilde{x}=0 . Assume that xXx\in X is outside the sphere. Then, f(x+0)f(0)+f(x).f(x+0)\leq f(0)+f(x). From the latter we get f(0)0f(0)\geq0 .

Thus, f(x)0f(x)\geq0 for all xX.x\in X.


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Comments

Assignment Expert
22.07.20, 23:02

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sana khyzer
22.07.20, 10:22

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