Closure of constant function 1 on the complex set

[victorvmotti]

TL;DR Summary

How can we find that the topological closure of the 1 constant function?

I'm watching this video to which discusses how to find the domain of the self-adjoint operator for momentum on a closed interval.



At moment 46:46 minutes above we consider the constant function 1

$$f:[0,2\pi] \to \mathbb{C}$$
$$f(x)=1$$

The question is that:

How can we show that the topological closure of the 1 function : $$\overline{\{1\}}$$ is all the complex constant functions on the domain above?

Put it another way, how can we show that a constant function can be a limit of a series of 1 function?

 

[martinbn]

The question is that:

How can we show that the topological closure of the 1 function : $$\overline{\{1\}}$$ is all the complex constant functions on the domain above?

Put it another way, how can we show that a constant function can be a limit of a series of 1 function?

This $\overline{\{1\}}$ means the smallest closed subspace that contains $\{1\}$, which is all constant functions.

 

[victorvmotti]

Sorry still not clear. I need a sort of easy to follow proof if possible. When we put a bar on a set, according to the video, we mean the topological closure, or that there exists a series with the limit equal to constant functions. But this is not clear for me.

 

[PeroK]

Sorry still not clear. I need a sort of easy to follow proof if possible. When we put a bar on a set, according to the video, we mean the topological closure, or that there exists a series with the limit equal to constant functions. But this is not clear for me.

It must be the topological closure of the subspace spanned by that function. The closure of a single function is just the function.

 

[victorvmotti]

Let me ask the question another way.

Consider the vector space of functions from a real interval to complex numbers.

If this vector space is seen as a topological space how can we apply the definition of the topological closure to show that the topological closure of the constant function 1, sending the interval to complex number 1, is the span of it.

In other words, why is that all the constant functions in this vector space is the topological closure of the constant function 1?

 

[PeroK]

In other words, why is that all the constant functions in this vector space is the topological closure of the constant function 1?

It isn't.

 

[victorvmotti]

So the lecturer is wrong?

I also thought that only the function 1 is the topological closure of the function 1.

And this claim confused me.

 

[PeroK]

So the lecturer is wrong?

I also thought that only the function 1 is the topological closure of the function 1.

And this claim confused me.

By "closure" he means "closure of the subspace spanned by the vector". Also, he says that the closure is the complement of the complement. Here he means "orthogonal complement of the orthogonal complement".

He's just using these terms as abbreviations.

 

[victorvmotti]

Thanks for the reassurance. Now it makes sense.

 

[martinbn]

Thanks for the reassurance. Now it makes sense.

If this takes so long to convince you, then how did you cope with the rest of the lecture?

 

[victorvmotti]

The rest was pretty clear. But this part what is written was not the same as what was said! Hence the confusion.
The closure of the "span" of the function 1 is all the constant functions. But as written on the blackboard it shows that as if the closure of the function 1 itself is all the constant functions.

 

[victorvmotti]

Here is the proof.

We show that $(\{1\}^\perp)^\perp =$ the set of constants functions on $[0,1]$.

One inclusion:

If $c$ is any constant and $\int_0^1 1 \cdot \bar{f} = 0$, then $\int_0^1 f \cdot \bar{c} = \overline{ c \int_0^1 1 \cdot \bar{f}} = 0$.

This shows all constants functions are contained in $(\{1\}^\perp)^\perp$.

The other inclusion:

Suppose that $g \in (\{1\}^\perp)^\perp$.

Now consider $h = g - \int_0^1 g \in \{1\}^\perp$

Note that $\langle h, h \rangle=\langle h, g-c \rangle=\langle h, g \rangle- \langle h,c \rangle$, where $c= \int_{0}^{1}g$

$\langle h, g \rangle=0$.

$\langle h, c \rangle = \langle g-c, c \rangle =0$.

$\langle h, h \rangle =0$.

$h=0$

$g=c$

This shows all functions contained in $(\{1\}^\perp)^\perp$ are constant functions.