I started studying sequences an series of functions and I found a problem which I don't quite understand. The problem is:

$f_n:(0,1)\rightarrow \mathbb{R}, f_n(x)=\frac{1}{nx+1}$, $n \geq 0$. Study the pointwise and uniform convergence of $f_n(x)$.

I showed that $f$ is pointwise convergent to $f(x)=0$ but I don't understand why is not uniform convergent.The solution in the book is:

$\underset{n \to \infty}{\lim}\hspace{1 mm} \underset{x\in (0,1)}{\sup} \Big | \frac{1 }{nx+1} \Big | =1$. Why? In another example I saw that he took $x=\frac{1}{n}$ to show that a sequence is not uniform convergent.

On series of functions: Can you give me some tips on how to show that a series is uniform convergent? I am confused because I don't know what theorem should I use or what result from the theory. Thank you.

  • $\begingroup$ You can compute the derivative, and find the maximum. $\endgroup$ – Itay4 Aug 9 '17 at 14:13

For $n>1$,

$$M_n=\sup_{x\in (0,1)}|f_n (x)-0|$$ $$=\sup_{x\in (0,1)}|f_n (x)|\ge |f_n (\frac {1}{n}) |$$

because $\frac {1}{n}\in (0,1) $.

but $$f_n (\frac {1}{n})=\frac {1}{2}$$ thus $$M_n\ge \frac {1}{2} $$ $$\implies \lim_{n\to+\infty}M_n\ge \frac {1}{2} $$ $$\implies \lim_{n\to+\infty}M_n \ne 0$$ hence, the convergence is not uniform at $(0,1) $.



When $f,g$ are bounded functions from $(0,1)$ to $\mathbb R,$ let $\|f-g\|=\sup \{|f(x)-g(x)|: x\in (0,1)\}. $

Uniform convergence of a sequence $(f_n)_n$ to $f$ is equivalent to $\lim_{n\to \infty}\|f-f_n\|=0.$

If $f(x)=0$ for all $x\in (0,1)$ then $\|f-f_n\|=\|f_n\|.$ So uniform convergence of $(f_n)_n$ to $0$ is equivalent to $\lim_{n\to \infty}\|f_n\|=0.$ But $\|f_n\|\geq |f_n(1/n)|=1/2,$ so convergence is not uniform.

Footnote: In fact, $\|f_n\|=1.$


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.