I'm reading Introductory functional analysis by Kreyszig and it discuss Fredholm alternative in the following way:
8.7-1 Definition (Fredholm alternative). A bounded linear operator $A:X \rightarrow X$ on a normed space $X$ is said to satisfy the Fredholm alternative if $A$ is such that either (I) or (II) holds:
- (I) The nonhomogeneous equations $Ax = y, A^\times f = g$ have solutions $x$ and $f$, respectively, for every given $y \in X, g \in X'$, the solutions being unique. The corresponding homogeneous equations $Ax = 0, A^\times f = 0$ have only the trivial solution $x = 0, f = 0$. ($A^\times$ is the adjoint operator of $A$ and $X'$ is the dual space of $X$.)
- (II) The homogeneous equations $Ax= 0, A^\times f = 0$ have the same number of linearly independent solutions $x_1, ..., x_n$ and $f_1, ..., f_n$, $n \geq 1$ resepctively. The nonhomogeneous equations $Ax = y, A^\times f = g$ are not solvable for all y and g, respectively; they have a solution if and only if y and g are such that $f_k(y) = 0, g(x_k) = 0$ for all $k = 1,2,..., n$.
What I'm confused about is on condition (II): Is there a contradiction between "The nonhomogeneous equations are not solvable for all $y$ and $g$" and "they have a solution iff ..."?
Any help will be appreciated
No, there is no contradiction. The equation $Ax=y$ is solvable if and only if $f_k(y)=0$ for all $k=1\dots n$.
The statement "The inhomogeneous equations are not solvable for all right-hand sides" is made precise in the second half of that long sentence "they have a solution if and only if...".