# Uniform convergence of some sequences of indicator functions in $\mathbb{R}$

We have the following sequences of functions: $$f_n(x)=1_{[\frac{1}{n+1}, \frac{1}{n}]}(x)$$ $$g_n(x)=\frac{1}{n}1_{[\frac{1}{n+1}, \frac{1}{n}]}(x)$$ $$h_n(x)=1_{[\frac{1}{(n+1)^2}, \frac{1}{n^2}]}(x)$$

All these functions are defined in $$[0,1]$$

Which of the following functions converge uniformly??

Now it is not difficult to see that all these sequences converge pointwise to $$0$$

Also $$\sup_{x \in [0,1]}|f_n(x)|=\sup_{x \in [0,1]}|h_n(x)|=1$$ proving that these sequences do not converge uniformly.

And $$\sup_{x \in [0,1]}|g_n(x)|= \frac{1}{n} \rightarrow 0$$

Therefore the only sequencce that converges uniformly is $$g_n$$

Is this correct or am i missing something?

Thank you in advance.

• As far as 'grading homework' goes, I guess it depends on the level of the detail that's expected. That said, it is certainly correct. Commented Aug 8, 2017 at 19:38
• I proved in detail the pointwise convergence ..but for uniform convergence i don't think that wants more detail..of course maybe i am wrong because this is a question that does not come from a homework especially a graded one Commented Aug 8, 2017 at 19:41

## 1 Answer

Just in case you might be interested;

Let $\varepsilon>0$ be given. Then there is a positive integer $N$ so that $N > \frac{1}{\varepsilon}$ (Archimedean Principle). So we now have that $$\sup\limits_{x\in[0, 1]} \left|g_n(x)\right|=\frac{1}{n} < \varepsilon \; \text{ whenever } \, n \geq N .$$ Therefore the sequence of functions $\{g_n\}_{n=1}^\infty$ converges uniformly on $[0,1]$ to the function $\lim_{n \to \infty} g_n(x)= 0$.

• OK...i already knew that but thanks anyway for your intension to help me..+1 from me..:) Commented Aug 9, 2017 at 11:13