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Consider function $f:\mathbb{R}^n \setminus \{0\} \times \mathbb{R}^n \rightarrow \mathbb{R}_{> 0}$ that is:

  • continuous in the first argument;

  • locally bounded in the second argument.

Consider a sequence $\{x_i\}_{i=1}^{\infty}$ such that $x_i \in \mathbb{R}^n$, $x_i \rightarrow x$.

Prove that

$$ \limsup_{i \rightarrow \infty} \ f(x,x_i) - f(x_i,x_i) \ = \ 0. $$

  • 0
    Can you write the condition locally bounded with "\forall", "$\exists$", etc...2012-03-09
  • 0
    I mean $\forall (x,y) \in \mathbb{R}^n \setminus \{0\} \times \mathbb{R}^n \ \exists \delta>0$ such that: $ |z-y|<\delta \Rightarrow |f(x,z)| for some $M>0$.2012-03-10
  • 0
    I guess $M$ depends on $x$ and $y$, right?2012-03-10
  • 0
    Yes: I mean that: http://en.wikipedia.org/wiki/Local_boundedness2012-03-10

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