Let $f:\mathbb R^+\longrightarrow \mathbb R$ continuous. Why $$\lim_{n\to \infty }\max_{1\leq j\leq 2^n}|f(jt2^{-n})-f((j-1)t2^{-n})|=0\ \ ?$$
Why $\lim_{n\to \infty }\max_{1\leq j\leq 2^n}|f(jt2^{-n})-f((j-1)t2^{-n}|=0$ where $f:\mathbb R\longrightarrow \mathbb R$ is continuous?
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real-analysis
probability
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0Because $f$ is uniformly continuous on $[0,t]$ for every fixed $t\in \mathbb{R}^+$. – 2017-01-04
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1What is $t$? And is the function $f$ defined on $\mathbb{R}$? or $\mathbb{R}^+$? – 2017-01-04
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0@DanielFischer: I understood Surb answer, but how uniform coninuity can gives you immediately the result ? – 2017-01-04
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0You are looking at $\lvert f(x) - f(y)\rvert$, where $\lvert x-y\rvert = t\cdot 2^{-n}$. By the uniform continuity, for every $\varepsilon > 0$ there is a $\delta > 0$ …, then for all large enough $n$ we have $t\cdot 2^{-n} < \delta$. – 2017-01-04
1 Answers
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First $jt2^{-n}\in [0,t]$ for all $j\in [0,2^n]$. By the way, since $f$ is continuous on $\mathbb R^+$, it's uniformly continuous on $[0,t]$. Now, let $j_n\in\{1,...,2^n\}$ s.t. $$|f(j_nt2^{-n})-f((j_n-1)t2^{-n})|=\max_{1\leq j\leq 2^n}|f(jt2^{-n})-f((j-1)t2^{-n})|.$$
Let $\varepsilon>0$ and let $\delta>0$ s.t. $|f(x)-f(y)|<\varepsilon$ whenever $|x-y|<\delta$. Let $N$ big enough to have $|jt2^{-n}-(j-1)t2^{-n}|<\delta$ whenever $n\geq N$. In particular, $$|f(j_nt2^{-n})-f((j_n-1)t2^{-n})|<\varepsilon,$$ whenever $n\geq N$, what prove the claim.