Let $I$ be an interval in $\mathbb{R}.$
For $k \in \mathbb{N}$, let $g_k, h_k \in L^2 (I)$ where $$\sum_{k \in \mathbb{N}} \left\Vert g_k \right\Vert_2^2 < \infty \quad \text{and} \quad \sum_{k \in \mathbb{N}} \left\Vert h_k \right\Vert_2^2$$
and $$\left\Vert g \right \Vert_2 = \bigg [\int_{I} \left|g(t)\right| ^2 dt \bigg ]^{\frac {1}{2}}$$.
Define $K : I \times I \rightarrow \mathbb{R}$ by $$K(x,y) = \sum_{n \in \mathbb{N}} g_n(x)h_n(y)$$ and $T : L^2(I) \rightarrow L^2(I)$ by $$Tf(x) = \int_I K(x,y)f(y)dy.$$
Suppose that there exists $C \in (0, \infty)$ such that $$\sum_{n \in \mathbb{N}} \left| \langle f,h_n \rangle \right| ^2 \ge C\left\Vert f \right\Vert ^2 \quad (\forall f \in L^2 (I))$$ and $$\sum_{j \in \mathbb{N}} \sum_{k \in \mathbb{N}} \left| \langle g_k, h_j \rangle \right|^2 < 1.$$ Show that $I - T : L^2(I) \rightarrow L^2 (I)$ is bijective. Hint: By the Fredholm Alternative, it suffices to show that $\mathcal{N}(I - T) = \{0 \}$.
I know that when the nullity is zero, then it is injective. By the Fredholm alternative, then it is injective if and only if it is surjective, thus, $I−T$ is bijective. I guess the help I need is by showing that $\mathcal{N}(I−T)=\{0\}.$