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.

  • $\begingroup$ 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. $\endgroup$ – Fimpellizieri Aug 8 '17 at 19:38
  • $\begingroup$ 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 $\endgroup$ – Marios Gretsas Aug 8 '17 at 19:41

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 \begin{equation} \sup\limits_{x\in[0, 1]} \left|g_n(x)\right|=\frac{1}{n} < \varepsilon \; \text{ whenever } \, n \geq N . \end{equation} 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 $.

  • 1
    $\begingroup$ OK...i already knew that but thanks anyway for your intension to help me..+1 from me..:) $\endgroup$ – Marios Gretsas Aug 9 '17 at 11:13

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