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Consider the space $C^1([0,1])$ and the function $d:C([0,1])\times C([0,1]) \to \mathbb R$ defined as $d(f,g)=|f(0)-g(0)|+sup_{x \in [0,1]}|f'(x)-g'(x)|$. Decide whether the metrics $d$ and $d_{\infty}$ are topologically equivalent in $C^1([0,1])$ (where $d_{\infty}=sup_{x \in [0,1]}|f(x)-g(x)|)$

My attempt at a solution:

If two metrics are topologically equivalent, then they have the same convergent sequences. Honestly, I couldn't do anything. I am trying to define a sequence of functions $\{f_n\}_{n \in \mathbb N}$ such that $f_n \to f$ in, for instance, $(C^1([0,1]),d)$ but $f_n \not \to f$ in $(C^1([0,1]),d_{\infty})$. Could it be this two metrics are topologically equivalent? If this is the case, how could I prove it? If not, I would appreciate any hint to find an adequate sequence of functions that works for what I am trying to prove.

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How about trying to figure out the open sets in each of the topologies? If there is some set that is open in one topology but not the other, then the topologies are different. – user99680 Dec 6 '13 at 4:01
Hmm, I am not sure if this set would work, but the set $U=\{f \in C^1[0,1]: f(x) \neq 0 \space \forall \space x \in [0,1]\}$ is open with $d_{\infty}$. Up to now, I couldn't prove/disprove is open with $d$. – user100106 Dec 6 '13 at 4:12
Let me check if it works. – user99680 Dec 6 '13 at 4:23
up vote 2 down vote accepted

Set $f_n(x) = \frac{1}{n} \sin(nx)$. It converges to zero in $(C^1([0,1]),d_{\infty})$ but it doesn't converge to zero in $(C^1([0,1]),d)$, because there is always a point $x$ such that $f'_n(x) = 1$.

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Thanks, I couldn't find a sequence that worked, this works perfectly. – user100106 Dec 6 '13 at 21:22

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