Functional Analysis

[visual echo]

Heyy, I really need help with this problem. My professor gave us a hint: choose the pre-hilbertspace C([0;1]) with the L^2 inner product. But I don‘t know what to do

[tulip flame]

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[quaint pendant]

what'd you try?

[quaint pendant]

also wdym by L2 inner product? I've only L2 followed by norm or metric

[visual echo]

$\int_0^1 f\cdot g$

[quaint pendant]

$\langle f\vert g\rangle =\int_0^1\overline{f}g?$

[mortal acorn]

Coffey2

Tony

[visual echo]

Yes

[quaint pendant]

Great

[visual echo]

We are in an R vectorspace tho

[quaint pendant]

did you try some U?

[visual echo]

Yes but I was stuck

I did

Uhh

$U={f : \int_0^{\frac{1}{2}}f=0}$

[mortal acorn]

Tony

[visual echo]

Ofc f is continuous

On [0;1]

It‘s closed (I checked). My idea is to show it‘s orthogonal complement is {0} and then take it‘s complement to get X which isn‘t equal to U obviously

I don‘t know how to show it‘s complement is {0}

Hmm

I could try proof by contradiction, no?

And then make use of the continuity of f maybe?

Lemme do it rq

[hard raptor]

Did you figure it out ?

[visual echo]

Yess

I showed that U is equal to the span of the indicator functions [0;1/2]

Thats pretty easy to show

Oh wait

I noticed a mistake in my proof

[hard raptor]

Yeah well this isn’t a continuous function

It’s not L^2[0,1] but C([0,1])

And I’m guessing you meant U^perp

[visual echo]

Uhh I fixed it

I think

Well

I showed U=span{indicatorfunction}^perp

Then perp both sides to get U^perp=span{indicator function}

[hard raptor]

But the indicator function isn’t continuous

[visual echo]

Except 0

[hard raptor]

it’s not in X

[visual echo]

So U^perp must be {0}, no?

Which means that (U^perp)^perp={0}^perp=X≠U

[hard raptor]

Yes, U^perp is {0}

[visual echo]

So the solution is correct, no?

[hard raptor]

i don’t understand your proof, like I said the indicator function isn’t continuous it’s not an element of your prehilbert space

you can’t use it here

[hard raptor]

You can prove this differently

[visual echo]

Uhh isn‘t C in L^2?

I mean Not literally

But like

And then argue that since U perp only consists of continuous functions, the only valid element in the span would 0

Since 0 is both in the span and continuous

Sooo, what do you think? Is it correct or are there any mistakes?

Oh shi

I see what you mean

Uhh

[hard raptor]

Actually, this argument does work but it assumes that you know stuff on L^2[0,1], ie that C[0,1] is dense it it for instance and that it’s indeed a Hilbert space

[visual echo]

Yes, I proved it before

[hard raptor]

you can also use a more standard argument

But ig this works

[visual echo]

Sure, I‘d love to see it

[hard raptor]

if g is orthogonal to U, then you can first show that g(x)=0 for any x in [1/2,1], by seeing that int_1/2^1 f(x) g(x) dx=0 for any continuous function f which is supported in [1/2,1]

[visual echo]

Proof by contradiction?

And then use the continuity to find a neighborhood where it‘s non zero?

[hard raptor]

well generally if $\mathcal{C}([a,b])$ has the $L^2$ inner product then ${f: f(a)=0}^{\perp}={0}$ and yeah you can show this by contradiction

[mortal acorn]

Rotator

[visual echo]

I see I see

[hard raptor]

So here you’ll have notably $g_{|[\frac{1}{2}1]} \in {f \in \mathcal{C}([\frac{1}{2},1]:f(\frac{1}{2})=0}^{\perp}$

[mortal acorn]

Rotator

[hard raptor]

got a bit mixed up but yeah

[visual echo]

I suppose you mean f(x)=0?

[hard raptor]

Because if f is continuous on [1/2,1] with f(1/2)=0 then you can extend it to [0,1] by simply setting f(x)=0 for x<1/2

and such an extension is supported in [1/2,1] by definition

[visual echo]

Ohh I see

[hard raptor]

now you can show that g is constant on [0,1/2]

Then you can conclude

[tulip flame]

@visual echo

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