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Suppose that $f:\mathbb R\longrightarrow\mathbb [0,+\infty[$ is an unlimited function. For every $n\in\mathbb N$ let's define the function $f_n(x)=\min\{f(x),n\}$, now my question is the following:

Does the sequence $\{f_n\}$ converge uniformly to $f$ in $\mathbb R$?

Intuitively the answer seems to be YES but in the effective calculations there are some indeterminate forms such as $\infty -\infty.$

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up vote 3 down vote accepted

The answer is definately NO: Take $f(x)=x$. The supremum of

$|f_n(x)-f(x)|=\max\{x-n, 0\}$

will always be unbounded.

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The answer is "no". The convergence is only pointwise. Define $f(x):=|x|$, then

$$|f_n(x)-f(x)| = \begin{cases} 0 & |x| \leq n \\ ||x|-n| & |x| \geq n \end{cases}$$

which shows that $f_n-f$ is unbounded (w.r.t. to supremum norm), in particular it does not converge to 0 (w.r.t. to supremum norm).

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Ok, so is the convergence pointwise for all $f$? – Dubious Dec 28 '12 at 18:19
No, pointwise for all $x \in \mathbb{R}$ (i.e. $|f_n(x)-f(x)| \to 0$ as $n \to \infty$)! – saz Dec 28 '12 at 18:21
Yes I mean that doesn't matter which is the function $f$. The convergence is always pointwise. – Dubious Dec 28 '12 at 18:27
Yes, the pointwise convergence holds always. But there are functions such that the convergence is even uniform (for example if $f$ is bounded, then $f_n = f$ for $n$ large enough, thus $f_n \to f$ uniformly). – saz Dec 28 '12 at 18:36


Let $f(x)=floor(x)$.

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In fact, all examples are counterexamples. – Hagen von Eitzen Dec 28 '12 at 18:27

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