## Uniform convergence

### Uniform convergence

Determine whether the sequence of functions $\displaystyle f_n(x)=e^{-x^2}\sin\frac xn$ converges uniformly on $\mathbb R$.

"Everything should be made as simple as possible, but not simpler." - Albert Einstein

### Re: Uniform convergence

How about this :

for any $\epsilon$ we can choose $x_0$ = $inf\{x : e^{-x^2} < \epsilon \}$

now we can choose $n$ large enough so that $sin(\frac{x}{n}) < \epsilon$ for all $|x| \leq x_0$

for any $\epsilon$ we can choose $x_0$ = $inf\{x : e^{-x^2} < \epsilon \}$

now we can choose $n$ large enough so that $sin(\frac{x}{n}) < \epsilon$ for all $|x| \leq x_0$

### Re: Uniform convergence

Does it not need to hold for all $x\in\mathbb R$?

"Everything should be made as simple as possible, but not simpler." - Albert Einstein

### Re: Uniform convergence

It does (right?)

For $|x| \leq x_0$ the $sin$ part is smaller than $\epsilon$ and for $|x|>x_0$ the $exp$ part is smaller than $\epsilon$. So the product is always smaller than $\epsilon$. (?)

For $|x| \leq x_0$ the $sin$ part is smaller than $\epsilon$ and for $|x|>x_0$ the $exp$ part is smaller than $\epsilon$. So the product is always smaller than $\epsilon$. (?)

### Re: Uniform convergence

That looks so easy compared to my one page long solution!

"Everything should be made as simple as possible, but not simpler." - Albert Einstein

### Re: Uniform convergence

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