Test whether or not $(f_n(x)) = \frac{nx}{n+x}$ on $I=[0,1]$ converges uniformly on $I=[0,1]$.

My attempt:

$\displaystyle \lim_{n \to \infty}f_n(x) = \lim_{n \to \infty}\frac{nx}{n+x} = \lim_{n \to \infty} \frac{x}{1 + \frac{x}{n}} = x$ for all $x \in I$.

Thus $(f_n)$ converges pointwise to $f(x) = x$ on $I$.

For uniform convergence, we must test the pointwise limit for uniform convergence:

$$||f_n - x|| = \sup\bigg\{ \bigg|\frac{nx}{n + x} -x\bigg| : x\in I \bigg\} = \sup \bigg\{\bigg| \frac{-x^2}{n+x}\bigg|: x \in [0,1] \bigg\} = \frac{1}{n+1}$$

Notice however, that $\frac{1}{n+1} \to 0$ as $n \to \infty$ and hence $(f_n)$ converges uniformly on $I$.

Is this correct?

  • $\begingroup$ It looks flawless to me. $\endgroup$ – Jack D'Aurizio May 14 '15 at 15:46

Thats very correct. Given any $\epsilon >0$ take $N(\epsilon) $ as $[\frac{1}{\epsilon}]+1$ and the definition follows.


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.