# Is there a subset which is bounded for one metric but not for the other?

We are given these two metrics on $C([0,1])$ (this space stands for the vector space of continuous functions from $[0,1]$ to $\mathbb{R}$): $d_{\infty}(f,g)= \sup \{|f(x)-g(x)|$ where $x \in [0,1] \}$ and $d_{1}(f,g)= \int_{0}^{1}|f(x)-g(x)|dx$.

Is there an example of a subset of $C([0,1])$ which is bounded for one of those metrics but not for the other? I reckon there isn't, since we're dealing with bounded functions here (the continuous funtions are defined on a closed interval), and so all subsets of $C([0,1])$ will be bounded for both metrics. Is this correct?

(Definition of boundedness: Let $(X,d)$ be a metric space and $A \subseteq X$. We say that $A$ is bounded if there exists an $M \in \mathbb{R}^{+}$ so that $d(x,y) \leq M$ for every $x,y \in A$.)

• Any subset which contains both $0$ and $f_n(x)=nx$ cannot be bounded over $d_{\infty}$, so what you said isn't correct. Commented Jun 12, 2017 at 17:06
• Observe $d_1(f,g) \leq d_{\infty}(f,g)$. So if a subset is bounded with respect $d_\infty$, it has to be bounded with respect to $d_1$. Commented Jun 12, 2017 at 17:07
• @SahibaArora But the converse of that statement is not true Commented Jun 12, 2017 at 17:13
• @ChristopherHalverson I think he/she was just suggesting that if you want to look for a counterexample then it must be that $d_{\infty}$-unbounded. Commented Jun 12, 2017 at 17:16
• @ChristopherHalverson Yes yes absolutely. user160738 had already covered that part so I didn't mention it. I was just putting light to the fact that while looking for a counterexample we need to find a subset which is bounded in $d_1$ but not in $d_{\infty}$. Commented Jun 12, 2017 at 17:17

Let $f_n$ be the piece-wise linear function which is $0$ at $0,\frac{1}{n}$ and $2n$ at $\frac{1}{2n}$. This class is bounded in $d_1$ as $d_1(f_m,f_n) \leq 2$ for all $m,n \in \mathbb{N}$, but is unbounded in the $d_{\infty}$ norm.