Is the Convergence Uniform?

[mathmari]

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:

 

[I like Serena]

Isn't the supremum $x\in \mathbb{R}$ the infinity?

Hey mathmari!

Yes, it is. (Nod)

 

[mathmari]

Yes, it is. (Nod)

So $f_n$ doesn't converge uniformly to$f^{\star}$, right? :unsure:

 

[I like Serena]

So $f_n$ doesn't converge uniformly to$f^{\star}$, right?

Indeed. :geek:

 

[HOI]

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!)