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This question concerns a problem from Walter Rudin's Functional Analysis textbook regarding boundedness in different topologies on $ C[0,1] $.

Let $ X = C[0,1] $ be the space of all complex-valued continuous functions on $[0,1]$. Consider the metric $ d(f,g) = \int_{0}^{1} \frac{|f(t) - g(t)|}{1 + |f(t) - g(t)|} dt. $ Additionally, define a topology $ \tau $ on $ X $ induced by the family of seminorms $ \mathcal{F} = \{ p_x : x \in [0,1] , \quad \text{where} \quad p_x(f) = |f(x)|. $

I am supposed to show that if a set $ E \subset X $ is bounded in $ (X,\tau) $, then it is also bounded in $ (X,\sigma) $. That is, the identity map $ \operatorname{Id} : (X, \tau) \to (X, \sigma) $ takes bounded sets to bounded sets. (Here $\sigma$ is the topology induced from the metric $d$ on $X$.)

My Attempt

Since $E$ is bounded in $ (X,\tau) $, for each $x \in [0,1] $, the set $ p_x(E) = \{ |f(x)| : f \in E \} $ is bounded in $ \mathbb{C} $. This means there exists $ M_x > 0 $ such that
$ |f(x)| \leq M_x, \quad \forall f \in E, \quad \forall x \in [0,1]. $
Thus, $E$ is a pointwise bounded family of continuous functions.

At this stage, I am unsure how to proceed further. Is there a way to leverage an Arzelà–Ascoli-type argument here? Or am I approaching this incorrectly?

Additional Concern

After reconsidering the problem, I noticed that for any $ f, g \in X $, the metric satisfies $ d(f,g) < 1 $. This suggests that $ (X,d) $ is a bounded metric space, meaning every subset of $X$ is automatically bounded with respect to $d$.

If this is the case, then the statement:

$ \text{If \( E \) is bounded in \( (X,\tau) \), then it is bounded in \( (X,d) \).} $

seems trivial. Am I missing something here? Is the intent of the problem simply to confirm that $ d $ is a bounded metric, or is there a deeper point being made?

Is the definition of a bounded topological vector space $(X,\sigma)$ not equivalent to boundedness of the metric $d$?

The question has been answered here, but the explanations are further increasing my cofusion. I would appreciate any insights or clarifications. Thanks in advance!

  • About the question you just deleted: Of course you did not claim that the space is $\mathbb R^2,.$ In my *humble* opinion: the best teaching method is to ask the students to assume that and prove that simple case first themselves. – Kurt G. Feb 18 '25 at 11:48
  • @KurtG. Yes, I totally agree. – Kishalay Sarkar Feb 18 '25 at 11:50

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