- Edit 2: As posteted I was missing crucial information.
"Let $H^1(0,1)$ be mappings $f:(0,1) \mapsto \mathbb{C}$ "
The right result can be proved using $$T:H^1(0,1)/H^1_0(0,1) \rightarrow \mathbb{C}^2; \quad f \mapsto T(f):= \lim(f(x_n),f(y_n))$$
Prove that $H^1(0,1)/H^1_0(0,1)$ is isomorphic to $\mathbb{C}^2$.
Here $H^1_0(0,1) := \overline{C^{\infty}_c(0,1)}$
I figured out a proof, but I'm quite uncertain if its right.
First define a mapping $T$. Let $x_n \rightarrow 1$ and $y_n \rightarrow 0$ then $$T:H^1(0,1)/H^1_0(0,1) \rightarrow \mathbb{C}^2; \quad f \mapsto T(f):= \lim(f(x_n),f(y_n),f'(x_n),f'(y_n))$$
- Edit 1 : Here $f'(x)$ denotes the weak derivative.
Now show : 1) $T$ is injective ; 2) $T$ is surjective ; 3) $T^{-1}$ is continuous
1)
Let $$f,g \in H^1(0,1)/H^1_0(0,1) \text{ with } T(f)=T(g)$$ then:
$$\lim f(x_n \to 1) = \lim g(x_n \to 1)
$$ $$\lim f(x_n \to 0) = \lim g(x_n \to 0)$$
For $h_0 := f-g $ we see that $h_0(x_n \to 1) = 0 = h_0(y_n \to 0)$ so $h_0 \in H^1_0(0,1)$.
Now $(g+h_0 = f) \Rightarrow (g \sim f)$ hence T is injective.
2) For $(a,b,c,d) \in \mathbb{C^2}$ define:
$$f(x) := \begin{cases}
cx+a &\text{for } x\in (0,\frac{1}{4})\\
\frac{1}{4}c+a+(2x-\frac{1}{4})(-\frac{1}{4}d+b-\frac{1} {4}c+a)&\text{for } x \in [\frac{1}{4},\frac{3}{4}]\\
dx+b-d &\text{for } x \in (\frac{3}{4},1)
\end{cases} $$
Then for the weak derivative we get:
$$f'(x) := \begin{cases}
c &\text{for } x\in (0,\frac{1}{4})\\
2(-\frac{1}{4}d+b-\frac{1} {4}c+a)&\text{for } x \in [\frac{1}{4},\frac{3}{4}]\\
d &\text{for } x \in (\frac{3}{4},1)
\end{cases} $$
Note: $f(y_n \to 0) = a, \quad f(x_n \to 1) = b, \quad f'(y_n \to 0) = c, \quad f'(x_n \to 1) = d $
From $ f \in L^2 $ and the existence of the weak derivative it follows that $f \in H^1(0,1)$
Concluding: $$\text{For every } z \in \mathbb{C}^2 \text{ we get a } f \in H^1(0,1)/H^1_0(0,1) \text{ with } T(f)=z$$
$T$ is surjective.
3) Take a look at the mapping defined in 2) and call it $T^{-1}$. $$T^{-1}: \mathbb{C}^2 \to H^1(0,1)/H^1_0(0,1) \quad (a,b,c,d) \mapsto f (x)$$ Now for $(a,b,c,d)_n \to (0,0,0,0)$ we get $$ \lim T^{-1}(a,b,c,d)_n = 0 \quad \text{for } x\in (0,1)$$ Hence $T^{-1}$ is continuous
As a bijective continuous mapping $T$ is an isomorphism and $H^1(0,1)/H^1_0(0,1)$ is isomorphic to $\mathbb{C}^2$.
Its a pretty unaesthetic proof and im not quite sure if it's right.
Mostly I wonder about the correctness of 3).
I would really appreciate some improvements :D
Thanks in advance.