Even more real analysis

[barren jolt]

Sorry guys… idk wtf this notation is- but ik what relative compactness is

[sick bay]

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[ember niche]

T is a linear map

that takes elements from C([a, b]) as inputs

and returns functions in C([a, b]) as outputs too

And for f in C([a, b]), Tf is a function

which formula is given to you there

See that C([a, b]) is a vector space, which origin is the function 0

which value is 0 everywhere in [a, b]

you can define the unit ball from that, I suppose from the metric induced by the L2 inner product

$\langle f, g \rangle = \int_{a}^{b} f(x)g(x)dx$

[proper star]

Rion

[ember niche]

So the unit ball is:
$$B = \left{ f \in C([a, b]) : \int_{a}^{b} f(x)^2 dx \leq 1 \right}$$

[proper star]

Rion

[barren jolt]

I chose the sup norm instead-

But yeah you were a great help

[barren jolt]

Then used Arzela-Ascoli to prove total-boundedness for T(B)

Now all that’s left is to prove that T(B) closure is compact

[ember niche]

I am not sure it helps but notice that T is continuous

For two simple and good reasons

1. It is linear
2. It is continuous at 0

So if you take a sequence g_n = T f_n of T(B), if f_n converges to f then g_n converges to T f

So it would in theory suffice to establish that any sequence of the unit ball has a convergent subsequence (if the unit ball is closed, then the subsequence limit is in the unit ball too)

Unfortunately you are somewhat doomed to failure given that C([a, b]) is not finite dimensional, therefore the unit ball is not compact

i.e. not every sequence admits a convergent subsequence within the unit ball

Never mind I spoke too soon, the Arzela-Ascoli gives you exactly that

[barren jolt]

Yeah Arzela-Ascoli is a nice theorem

[ember niche]

I mean, let's exploit it since we can

Let $(g_n)$ a sequence in $\overline{T(B)}$. By definition, for every $n$, there exists $f_n \in C([a, b])$ such that $g_n = T f_n$.

[proper star]

Rion

[barren jolt]

Huh true

So if i prove that the original ball is compact

With sup borm

[ember niche]

Actually I was about to write some bs but I changed my mind

It is true that $f_n$ is uniformly bounded by 1, by definition

[proper star]

Rion

[ember niche]

but it is not true that $f_n$ is uniformly equicontinuous

[proper star]

Rion

[ember niche]

Just because it lies in the unit ball

Consider $f_n : x \mapsto k_n (x - a)$ if that's less than $1$, and $1$ otherwise, where $k_n = n$

[proper star]

Rion

[ember niche]

It's not compact though

the unit ball is compact if and only if the ambient space is of finite dimension

[barren jolt]

*it’s closure

Bruh extra apostrophe

[barren jolt]

But a suitable metric could cause that, right?

[ember niche]

Nope

[barren jolt]

I did prove compactness for the space in the third question though (the question i posted previously)

I have to prove that it’s closure is compact… I still think that a suitable metric would decrease the amount of dimensons-

Dimensions

[ember niche]

https://math.stackexchange.com/questions/287360/is-it-true-that-the-unit-ball-is-compact-in-a-normed-linear-space-iff-the-space

[barren jolt]

I visualized the “infinite dimension function space” as a line-

[ember niche]

but it's not a line

one vector does not suffice to span it

[barren jolt]

Arzela Ascoli suffices to prove total boundedness for it

and the closure is always closed

so um I proved it by accident