I'm learning about uniform convergence. For example, consider a function sequence

$$f_n: \mathbb{R} \rightarrow \mathbb{R}, f_n(x) = n\sin(\frac{x}{n})$$

This sequence converges pointwise to $ f(x) = x $, but on $ \mathbb{R} $ the convergence is not uniform (one can look at $ f(\frac{\pi}{2n}) = n$ ).

I'm supposed to prove that on any closed interval $ [a,b] \subset \mathbb{R} $ the convergence is uniform.

To prove that I need to show that the supremum of

$$ | x - n\sin(\frac{x}{n}) | $$

converges to $ 0 $ as $n \rightarrow \infty $.

My take is this: if $ x \in [a,b] \subset \mathbb{R} $, then by Weierstrass theorem a continuous function on closed interval attains its maximum value, so there exists $ x_1 \in [a,b] $ that for every $x$

$$ | x - n\sin(\frac{x}{n}) | \leq | x_1 - n\sin(\frac{x_1}{n}) |$$

and then it's easy to see that as $n \rightarrow \infty $ the RHS also converges to $0$. But RHS is the supremum which converges to $0$, so the convergence is uniform on any $ [a,b] \subset \mathbb{R} $.

Is this method/trick correct? It seems like if it is, then proving almost uniform convergence can sometimes be easy just by invoking the Weierstrass theorem: suspiciously easy, that's why I'm asking.

  • 2
    $\begingroup$ The value of $x_1$ depends on $n$. $\endgroup$
    – TSF
    Apr 21 '19 at 11:57

Actually, you don't have $f_n\left(\frac\pi{2n}\right)=n$. You can prove that the convergence is not uniform on $\mathbb R$ usinge the fact that $f_n\left(\frac{\pi n}2\right)=n$ for each natural $n$.

And your proof is wrong, since $x_1$ depends upon $n$.

I suggest that you define $g_n(x)=f_n(x)-x=n\sin\left(\frac xn\right)-x$ and then you prove that it converges uniformly to the null function on $[a,b]$. Use the fact that $g_n'(x)=\cos\left(\frac xn\right)-1$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.