Why is this false:
We have given $V = \mathbb{R}[x]_{\leq 3}$ as the vectorspace of the polynomials with variable $x$ with a maximum degree of $3$ . Also we do have the linear forms $\lambda_1, \lambda_2 , \lambda_3, \lambda_4 \in V^{*}$ (where $V^*$ is the dual space of $V$) with
$\lambda_1(f) = f(1), \lambda_2(f) = f'(1), \lambda_3(f) = f(-1)$ and $\lambda_4(f) = f'(-1)$.
I now want to show that
$\lambda_1, \lambda_2 , \lambda_3, \lambda_4$ is a basis of $V^*$.
What I thougt I should do is to show that for a basis $B = (1,x,x^2,x^3) = (x_1,...,x_4)$ of $\mathbb{R}[x]_{\leq 3}$ follows
$\lambda_i(x_j) = \delta_{i j}$
But thats actually not the case. I do have the solution for this but Im wondering why thats not the rigth thing to show here.
It is indeed true that the biorthogonality condition
$$\lambda_i(x_j) = d_{ij}$$ should be fulfilled, if the $\lambda_i$ form a dual basis of the $x_j$. However, you cannot just take any basis of the dual space $V^*$ and any basis of $V$ and claim that the above condition holds true. Your $\lambda_i$ form a basis of $V^*$, but they are not the dual basis to your chosen $B = (1,x,x^2,x^3)$.
If you want to find the basis $B$ of $V$, such that the orthogonality condition holds for your given dual basis $\{\lambda_i\}_{i=1}^4$, you have to write your $\lambda_i$ in vector form, put them row-by-row into a matrix and then invert this matrix as follows:
Let $$A = \begin{pmatrix} 1 & 1 & 1 & 1\\ 0 & 1 & 2 & 3\\ 1 & -1 & 1 & -1\\ 0 & 1 & -2 & 3 \end{pmatrix}$$
represent your dual basis. You can get the $i$-th row of $A$ by just applying $\lambda_i$ to $(1,x,x^2,x^3)$ each. So for example the first row is just $$(\lambda_1(1),\lambda_1(x),\lambda_1(x^2),\lambda_1(x^3))$$Inverting this gives $$A^{-1} = \begin{pmatrix} 0.5 & -0.25 & 0.5 & 0.25\\ 0.75 & -0.25 & -0.75 & -0.25\\ 0 & 0.25 & 0 & -0.25\\ -0.25 & 0.25 & 0.25 & 0.25 \end{pmatrix}$$
Interpreting the collumns of $A^{-1}$ as functions will give you the desired result. For example, lets take the first column which represents $f(x) = 0.5+0.75x-0.25x^3$. Then plugging in will give you $f(1) = 1$, but also $f'(1)= 0 $ etc.