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I think the right sequence is $f_n(t)=|t-\frac{1}{2}|^{(n+1)/n}$ but I can't manage to prove it's Cauchy... (when I look at graph of $f_n$s it seems obvious, but I need it formal way.

luka5z
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    Try finding $\lVert f_n - f\rVert_\infty$, where $f(t) = \lvert t-\frac{1}{2}\rvert$. That may be easier. – Daniel Fischer Apr 22 '14 at 09:27
  • But I know $f_n(t)$ is Cauchy, so one must be able to prove it's really Cauchy. SO you propose alternative approach? Let $f_n$ remain unknown and find one that satisfies $\forall \epsilon>0 \dots$ – luka5z Apr 22 '14 at 09:31
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    What Daniel Fischer means is that $f_n$ converges to $f$ in $C^0([0,1])$, so $f_n$ is necessarily Cauchy in $C^1([0,1])$. – Seirios Apr 22 '14 at 09:39
  • Because every convergent sequence is Cauchy sequence, yes? – luka5z Apr 22 '14 at 09:46
  • @DanielFischer I'm a bit puzzled -- the $f_n$ in the question involve an absolute value which is not differentiable at $0$. Is $f_n(t) = |t-1/2|^{(n+1)/n}$ really in $C^1$? – Rudy the Reindeer Apr 22 '14 at 14:48
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    @MattN. Yes, $$\frac{d}{dx}\lvert x\rvert^{1+\varepsilon} = (1+\varepsilon)\operatorname{sign}(x)\cdot \lvert x\rvert^\varepsilon$$ is continuous for $\varepsilon > 0$. – Daniel Fischer Apr 22 '14 at 14:55

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