- #1
mathmari
Gold Member
MHB
- 5,049
- 7
Hey! :giggle:
We have the sequence of functions $$f_n=\sin (x)-\frac{nx}{1+n^2}$$ I want to check the pointwise andthe uniform convergence.
We have that $$f^{\star}(x)=\lim_{n\rightarrow \infty}f_n(x)=\lim_{n\rightarrow \infty}\left (\sin (x)-\frac{nx}{1+n^2}\right )=\sin(x)$$ So $f_n(x)$ converges pointwise to$f^{\star}=\sin(x)$.
We have that $$\left |f_n(x)-f^{\star}(x)\right |=\left |\sin (x)-\frac{nx}{1+n^2}-\sin(x)\right |=\left |-\frac{nx}{1+n^2}\right |$$ We have to calculate first the supremum for $x\in \mathbb{R}$ and then the limit for $n\rightarrow \infty$.
Isn't the supremum $x\in \mathbb{R}$ the infinity? :unsure:
We have the sequence of functions $$f_n=\sin (x)-\frac{nx}{1+n^2}$$ I want to check the pointwise andthe uniform convergence.
We have that $$f^{\star}(x)=\lim_{n\rightarrow \infty}f_n(x)=\lim_{n\rightarrow \infty}\left (\sin (x)-\frac{nx}{1+n^2}\right )=\sin(x)$$ So $f_n(x)$ converges pointwise to$f^{\star}=\sin(x)$.
We have that $$\left |f_n(x)-f^{\star}(x)\right |=\left |\sin (x)-\frac{nx}{1+n^2}-\sin(x)\right |=\left |-\frac{nx}{1+n^2}\right |$$ We have to calculate first the supremum for $x\in \mathbb{R}$ and then the limit for $n\rightarrow \infty$.
Isn't the supremum $x\in \mathbb{R}$ the infinity? :unsure: