2

These days I am reading topology.

I came to know about a very beautiful result which says that all the norms on the finite dimensional vector space are equivalent. Using this it was very easy to see that on $GL(n,R)$ all the topologies induced by different norm are equivalent.

Now I am trying to analyze $S=C([a,b])$, set of all continuous real valued function from a compact set $[a,b]$ to $\mathbb{R}$. So far I have seen these three metric in literature.

$X_1: d(f,g)=\sup|f-g|$

$X_2: d(f,g)=\int_{0}^{1}|f-g|dt$

$X_3: d(f,g))=(\int_{0}^{1}|f-g|^2)^{1\over 2}$

Now I want to compare these topologies. I want to deduce which topologies is finer, which is coarser.

To be honest, I don't have any intuition on how to start visualizing open balls in the space.

I would be glad if somebody can show me the path. Thanks in advance.

Shweta Aggrawal
  • 5,659
  • 1
    Given that $C([a,b])$ is infinite-dimensional, visualization is going to be hard. What can help build intuition is by comparing them to the $\ell^\infty$, $\ell^1$, and $\ell^2$ metrics on $\mathbb R^n$, respectively. – Aweygan Jan 15 '19 at 14:24
  • After writing an answer I noticed this question https://math.stackexchange.com/questions/1529156/uniform-convergence-implies-l2-convergence-and-l2-convergence-implies-l1 (It's not exactly a duplicate but with a little effort it gives you what you need). – Yanko Jan 15 '19 at 14:33

1 Answers1

2

It's hard to visualize because the vector space $C([a,b])$ is infinite dimensional. But you can use these inequalities:

First $$\int_a^b |f-g|dx \leq (b-a)\sup|f-g|$$

Which means that $\sup|f-g|$ is small implies that $\int_a^b |f-g|dx$ is small.

Similarly $$(\int_a^b |f-g|^2dx )^\frac{1}{2} \leq ((b-a)^2\sup|f-g|^2)^{\frac{1}{2}} = (b-a)\sup|f-g|$$

These two should teach you about the relation between $X_2,X_1$ and $X_3,X_1$. It is left to study the relation between $X_2,X_3$. For this you need a less trivial inequality called the holder inequality. It says that

$$\int_a^b |f(x)-g(x)|dx \leq (b-a)^{1/2}\cdot (\int_a^b |f(x)-g(x)|^{2}dx)^\frac{1}{2}$$

Which means that if the $X_3$'s distance is small then the $X_2$'s distance is small.

Yanko
  • 14,341
  • After writing an answer I noticed this question https://math.stackexchange.com/questions/1529156/uniform-convergence-implies-l2-convergence-and-l2-convergence-implies-l1 (It's not exactly a duplicate but with a little effort it gives you what you need). – Yanko Jan 15 '19 at 14:33
  • thanks a lot... – Shweta Aggrawal Jan 15 '19 at 14:40