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Prove that the following function converges pointwise: $f_n: [0,1]\to \mathbb{R}$ with $f_n(x)=\dfrac{\sin{nx}}{\sqrt{n}}$

My proof

$f(x)=0$

Choose $x_0$ in $[0,1]$ randomly, fix $n^*=\dfrac{1}{\sqrt{\epsilon}}$, then if $n>n^*$, we get:

$\left\lvert f_n(x_0)-f(x_0)\right\rvert=\left\lvert\dfrac{\sin{\left(nx_0\right)}}{\sqrt{n}}\right\rvert\leq \dfrac{1}{\sqrt{n}} < \dfrac{1}{\sqrt{n^*}}=\epsilon$

So the function converges.

Does this hold?

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  • $\begingroup$ It seems ok. Just be careful that, as it's been defined, $n^*$ needs not be integer (though you don't need it to be). And, perhaps, you may want to evaluate it at "$x_0$", since you took the time to pick it. But "$x$" works just fine for the name of the point. $\endgroup$
    – user228113
    Feb 18, 2016 at 15:35
  • $\begingroup$ @G.Sassatelli, you're absolutely right about evaluating at $x_0$ since I introduced it for that particular job. All the proofs I've seen use the squeeze theorem which made me doubt my own proof. Thanks for your feedback! $\endgroup$
    – Matthew
    Feb 18, 2016 at 15:50
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    $\begingroup$ You picked the wrong $n^{\ast}$. You want $\dfrac{1}{\sqrt{n^{\ast}}} \leqslant \epsilon$, so you must take $n^{\ast} \geqslant \dfrac{1}{\epsilon^2}$, not $\sqrt{\epsilon}$. $\endgroup$
    – Daniel Fischer
    Feb 18, 2016 at 15:59
  • $\begingroup$ Whooops, that was quite sloppy of me not to notice. $\endgroup$
    – user228113
    Feb 18, 2016 at 16:18

2 Answers 2

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As pointed out in the comments:

  • The choice of $n^*$ has two problems. First it should be an integer, and second this integer should be greater than $1/\varepsilon^2$ instead of $1/\sqrt\varepsilon$.
  • Actually you do not choose randomly an $x_0$. You fix $x_0$ in the unit interval and work with it. There is nothing random here.
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Hint: $|\sin(nx)|\leq 1$, for each $n\in\mathbb{N}$.

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