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Given that we have sequence of piece-wise functions $S_{n}$ on [0,2] given by $ \frac{1}{n}x+x^{2}$ for ,$ 0\leq x \leq 1$ and $\frac{1}{2n}$ for , $1<x \leq 2$

where sup metric distance between functions $ d_{s}(h,g) = \underset{x \in {[0,2]}}{sup} | f(x) -g(x) |$

How can we find the expression for the sup distance between any two functions for any arbitrary n under sup metric.

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    $\begingroup$ What did you try? Find the extemum of the function $f-g$ by looking at the points where $f'-g'=0$ ... $\endgroup$ – LL 3.14 May 3 '20 at 7:29
  • $\begingroup$ I am trying to solve for uniform convergence of this sequence of functions. So, i know its true that the function converge in point-wise, but to show the uniform convergence in this case, it is intuitive that the max distance between any two consecutive functions is at x=1, but i am not sure whether it is true for all two arbitrary functions. In addition, if i can have expression for max distance between any two functions from this sequence for arbitrary n, then i can use the definition of uniform convergence to prove. $\endgroup$ – Muhammad May 3 '20 at 7:33
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Hint

$$\sup_{x\in[0,2]} |f_n(x)-f_m(x)|=\max\Bigg\{\sup_{x\in[0,1]} |f_n(x)-f_m(x)|,\sup_{x\in(1,2]} |f_n(x)-f_m(x)|\Bigg\}$$where $f_n$ and $f_m$ are two arbitrary functions for $m,n\in\Bbb N$.

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  • $\begingroup$ Thank you and, i have used this and now i have to show $| \frac{1}{n}-\frac{1}{m}| <\varepsilon $ for $\forall n,m>N \in \mathbb{N}$ . Do you have any hint for that. $\endgroup$ – Muhammad May 3 '20 at 8:14
  • $\begingroup$ You're welcome. This proof follows very easily. Let $n>m$ by symmetry and use $$|{1\over n}-{1\over m}|={1\over m}-{1\over n}<{1\over m}<\epsilon$$ $\endgroup$ – Mostafa Ayaz May 3 '20 at 8:42
  • $\begingroup$ Yes this also works. I used the fact that n,m $\geq$ r where r >N so $\frac{1}{n} < \frac{1}{N} $ and because $\frac{1}{m}>0$ therefore, $\frac{1}{n}-\frac{1}{m} < \frac{1}{N}$ and by archimedian property such N exist such that $N > {1/\varepsilon}$. Is this reasoning correct.? $\endgroup$ – Muhammad May 3 '20 at 10:00

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