Point to point convergence / Uniform convergence
- Tolaso J Kos
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Point to point convergence / Uniform convergence
Does the sequence of functions \( \displaystyle f_n(x)=\frac{x^{2n}}{1+x^{2n}}, \; n=1, 2, \dots , x \in \mathbb{R} \) converge point to point?
Is the convergence uniform?
Is the convergence uniform?
Imagination is much more important than knowledge.
- Grigorios Kostakos
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Re: Point to point convergence / Uniform convergence
Easy to check that \[f_{n}(x)\xrightarrow{pointwise}f(x)=
\begin{cases}
1\,, & x\in(-\infty,-1)\cup(1,+\infty)\\
0\,, & x\in(-1,1)\\
\frac{1}{2}\,, & x=\pm1
\end{cases}\,.\] Because the functions \(f_n(x)\) are continuous and the function \(f\) is not continuous, the sequence \(\{{f_n}\}_{n\in\mathbb{N}}\) does not converges uniformly to \(f\).
\begin{cases}
1\,, & x\in(-\infty,-1)\cup(1,+\infty)\\
0\,, & x\in(-1,1)\\
\frac{1}{2}\,, & x=\pm1
\end{cases}\,.\] Because the functions \(f_n(x)\) are continuous and the function \(f\) is not continuous, the sequence \(\{{f_n}\}_{n\in\mathbb{N}}\) does not converges uniformly to \(f\).
Grigorios Kostakos
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