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I wonder if this is true: Let $(f_n)$ be a sequence of real-valued functions defined on a set $S\subset\mathbb{R}$. Assume that the function $f(x)$ is continuous and $\lim_{n\to\infty}f_n(x)=f(x)$. Does this imply that $f_n(x)$ is uniform convergent? The other way around it seems to be true, according to my book:
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No. Here is a counterexample. $f_n(x) = \frac{x}{n}$ converges pointwise (but not uniformly) to the constant function $f(x) = 0$ on $\Bbb R$, which is continuous. |
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No it does not. As a counterexample, consider the function $f:(0,1)\to\Bbb R$ defined by $f(x)=1/x$. Clearly this is continuous; now consider the sequence of functions $$f_n(x)=\begin{cases} n &: x<1/n\\ 1/x &: 1/n\leq x<1\end{cases}.$$This is a standard example where I've merely modified the domain in order to preserve continuity. As an aside, this example also shows that monotone convergence (i.e., $f_m\leq f_n$ for $m<n$) need not imply uniform convergence if the domain is bounded but not compact (an idea that Dominic Michaelis mentions in his answer). |
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The result is not true in general. Let $f_n\colon[0,1]\to\mathbb{R}$ be defined by $f_n(x)=4\,n\,x(1-n\,x)$ if $0\le x\le1/n$, $f_n(x)=0$ if $1/n<x\le1$. Then $f_n$ converges pointwise but not uniformly to the continuous function $f(x)=0$. On the other hand, you should check Dini's theorem. |
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As it is mentioned it is not true. But if you add some more Assumptions like $S$ is compact and $f_n(x)\geq f_m(x)$ for $n>m$ (I think you don't need that $f_n$ is continuous), than you can show, that the convergence is uniform. |
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