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Let us consider $C[0,1]$, the space of continuous functions $f\colon [0,1] \to \mathbb{R}$. It comes usually with the metric of the maximum, or of the supremum, $d_{L^{\infty}}$. Each element $f$ in $C[0,1]$ corresponds uniquely to a compact subset of $\mathbb{R}^2$, by taking the graph of $f$. Thus, we can consider the Hausdorff metric $\rho$ induced on $C[0,1]$.

My question is: "Is the Hausdorff metric $\rho$ on $C[0,1]$ equivalent to the (usual) $L^{\infty}$-metric?". (equivalent metrics for me means the underlying topologies are the same).

I understand that $\rho \leq d_{L^{\infty}}$. Also, the two metrics are not uniformly equivalent, indeed there exists no universal constant $C$ such that $d_{L^{\infty}} \leq C \cdot \rho$. To see this, just take the sequence of functions $f_n(x)= nx$ and $g_n(x)= nx+1$. The infinity distance of $f_n$ to $g_n$ is constantly equal to $1$, but the Hausdorff distance equals $1/n$.

(I am sorry if the question is too trivial/elementary, feel free to cancel it).

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Yes, this is true.

You need to show that for any $\varepsilon>0$ and any function $f$ in $C[0,1]$ there is $\delta>0$ such that $\varepsilon$-neighborhood in one metric contains $\delta$-neighborhood in the other metric and the other way around.

One way you know already; it follows since $\rho\le d_{L^\infty}$.

The other way it follows since any continuous function $[0,1]$ is uniformly continuous. Indeed, it is easy to see that if $\omega$ is a modulus of continuity of $f$ and $\rho(f,g)<\varepsilon$ then $d_{L^\infty}(f,g)<\omega(\varepsilon)+\varepsilon$.

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  • $\begingroup$ Thanks for your answer. Can you elaborate? I do not see your point. $\endgroup$ – calc Nov 21 '13 at 19:39
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    $\begingroup$ Ok, I think I see the point. For any $\epsilon$ infinity-ball $B$ around $f$, you want a $\delta$ $\rho$ ball around $f$ that is inside of B. You take as $\delta$ the $\delta_1/2$, where $\delta_1$ is the output of uniform continuity of $f$ for the input $\epsilon$. (for simplicity, here I am using the $1$-metric on $\mathbb{R}^2$, otherwise the formula for $\delta$ in terms of $\delta_1$ would be more complicated.) $\endgroup$ – calc Nov 21 '13 at 20:51
  • $\begingroup$ Thanks also for the revision, I think it is more clear now. $\endgroup$ – calc Nov 22 '13 at 16:37

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