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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

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

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Can you write the condition locally bounded with "\forall", "$\exists$", etc... – Davide Giraudo Mar 9 '12 at 18:20
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)|<M$ for some $M>0$. – Adam Mar 10 '12 at 7:04
I guess $M$ depends on $x$ and $y$, right? – Davide Giraudo Mar 10 '12 at 10:48
Yes: I mean that: – Adam Mar 10 '12 at 17:53

I believe the following is a counter-example.

To keep things simple, let $n=1$. Consider the sequence $x_n = 1-\frac{1}{2^n}$. This sequence converges to $x=1$. Now define a function $f:\mathbb{R}\setminus\lbrace0\rbrace\times\mathbb{R}\to\mathbb R_{>0}$ as follows: $$f(t,y):=\begin{cases}1&&\textrm{if }y\geq1\\1 && \textrm{if }t\leq y<1\\1+\frac{t-x_n}{x-x_n}&&\textrm{if }y<t<1\\2&&\textrm{if }y<1\leq t\end{cases}$$ This function is bounded (it takes values between $1$ and $2$), so it is also locally bounded in the second argument. It is also continuous in the first argument, since it is piecewise linear for each fixed $y$ and the boundary values (or limits) agree.

But, $f(x,x_n)-f(x_n,x_n)=2-1=1$ for each $n\in\mathbb{N}$, so $\lim_{n\to\infty}(f(x,x_n)-f(x_n,x_n))=1$, which means $\limsup$ is also equal to $1\neq0$.

In case you want a counter-example for $n\in\mathbb{N}$, a similar construction should work. The point here is that you can define a continuous function $f_y:\mathbb{R}^n\setminus\lbrace0\rbrace\to\mathbb{R}_{>0}$ pretty much independently for each $y\in\mathbb R^n$ and then just put them together by $f(t,y) = f_y(t)$ as long as you make sure they stay sufficiently bounded. This independence can then enable some pretty wild behaviour.

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