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Is the convergence uniformly?

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,863
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:
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,734

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,863

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,734

Country Boy

Well-known member
MHB Math Helper
Jan 30, 2018
881
Comment on Grammer: "uniformly" is an adverb and so modifies to verbs, adjectives, and other adverbs. Here "converge" is a noun and so requires the adjective "uniform".

One can ask "Does this converge uniformly?" or "Is this convergence uniform?" but not "Is this convergence uniformly".

(Yes, I realize this was probably just a typo but I couldn't help myself!)