# Supremum distance between any two functions in the sequence

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

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.

• What did you try? Find the extemum of the function $f-g$ by looking at the points where $f'-g'=0$ ... – LL 3.14 May 3 '20 at 7:29
• 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. – Muhammad May 3 '20 at 7:33

$$\sup_{x\in[0,2]} |f_n(x)-f_m(x)|=\max\Bigg\{\sup_{x\in[0,1]} |f_n(x)-f_m(x)|,\sup_{x\in(1,2]} |f_n(x)-f_m(x)|\Bigg\}$$where $$f_n$$ and $$f_m$$ are two arbitrary functions for $$m,n\in\Bbb N$$.
• 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. – Muhammad May 3 '20 at 8:14
• You're welcome. This proof follows very easily. Let $n>m$ by symmetry and use $$|{1\over n}-{1\over m}|={1\over m}-{1\over n}<{1\over m}<\epsilon$$ – Mostafa Ayaz May 3 '20 at 8:42
• 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.? – Muhammad May 3 '20 at 10:00