# Proof of pointwise convergence

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?

• 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.
– user228113
Feb 18, 2016 at 15:35
• @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! Feb 18, 2016 at 15:50
• 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 $\sqrt{\epsilon}$. Feb 18, 2016 at 15:59
• Whooops, that was quite sloppy of me not to notice.
– user228113
Feb 18, 2016 at 16:18

• 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.
Hint: $|\sin(nx)|\leq 1$, for each $n\in\mathbb{N}$.