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As the title says, I am looking for a sequence of function which is uniformly convergent on all compact sets of $\mathbb R$ but not on $\mathbb R$.

I thought $f_n(x) = \frac{x}{n}$ is such a function since for any x in a bounded and closed subset of $\mathbb R$. $\sup(f_n(x)-f(x)) \to 0$ as $n\to\infty$. But since $\mathbb R$ is unbounded $x$ can get infinitely large thus the function sequence does not uniformly converge on $\mathbb{R}$. I wanted to check if my understanding is correct. Thank you

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    $\begingroup$ Yes - good example. $\endgroup$ – RRL Jan 6 at 8:38
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    $\begingroup$ Check my edits to improve your MathJax skills. $\endgroup$ – RRL Jan 6 at 8:41
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    $\begingroup$ Here is another example of what you are looking for where it is a little more difficult to prove uniform convergence on the compact intervals. $\endgroup$ – RRL Jan 6 at 8:52
  • $\begingroup$ @RRL thanks a lot $\endgroup$ – Kaan Yolsever Jan 6 at 8:59
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    $\begingroup$ Another, more difficult example, is $f_n(x)=\sum_{j=0}^nx^j/j!$ with $f(x)=e^x.$ A theorem in analysis is that if a power series in $x$ converges at every $x$ then the convergence is uniform on any compact set. But $f_n$ is a polynomial and $e^x-p(x)\to \infty$ as $x\to \infty$ for any polynomial $p.$ $\endgroup$ – DanielWainfleet Jan 30 at 23:21
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As explained above in the post, the answer is {$f_n(x) = \frac{x}{n}; n \geq 1$ and $x \in \mathbb R $}

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