If sequence of functions $f_{n}$ in $C_C(\mathbb{R})$ converges pointwise to the functions $e^{-x^2}$, then can we conclude $f_n$ converge uniformly.
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This may be of help: https://math.stackexchange.com/q/597765/353004 – astudentofmaths Jun 18 '18 at 21:45
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No. Consider the sequence $$ f_n(x) = (e^{-x^2}+\chi_{[n,n+1]}), $$ where $\chi$ is the indicator function.
Notice that for any $x\in \mathbb{R}$, once $n>x$ then $f_n(x)=e^{-x^2}$, so $f_n$ converges pointwise.
On the other hand $\|f_n-e^{-x^2}\|_\infty = \|\chi_{[n,n+1]}\|_\infty = 1$ for all $n$.
I now realize you want functions in $C_C(\mathbb{R})$, but hopefully you see how to modify this example slightly to that end
Matt
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@mathlover Rather than take $\chi_{[n,n+1]}$, take your favorite bump function in $C_c(\mathbb R)$ and add translates of it to obtain the $f_n$. – Aweygan Jun 22 '18 at 20:35
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@Aweygan we can take $f_n(x)=e^{-x^2}\chi_{[0,\frac{1}{n}]}(x)$. Right? Then $||f_n-e^{-x^2}||_{\infty}\to 0$ as $n\to \infty$. Then $(f_n)$ uniformly converges to $e^{-x^2}$. Am I right? – Aritra Jun 26 '22 at 08:58
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@Aritra That's not continuous, and $|f_n-e^{-x^2}|_\infty=1$ for all $n$. What are you aiming for? – Aweygan Jun 27 '22 at 13:58
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@Aweygan I am trying to find a sequence of function $f_n\in C_c(\mathbb R)$ such that $(f_n)$ convergers uniformly to $e^{-x^2}$. – Aritra Jun 27 '22 at 15:40