Bound on inverse operator

Define $X = {C^{2, \alpha}}(U \times [0,T])$ and $Y = {C^{0, \alpha}}(U \times [0,T])$ where $U$ is some real interval.

Let $F:X \to Y$ be a map. Let $DF(g):X \to Y$ be a bounded linear operator for each $$g \in \{C^{2, \alpha}(U \times [0,T]) \} \cap \{g : |g(x,t)| < w\}.$$ Note that $DF(g)$ is the Frechet derivative of F at point $g$.

Suppose we have $$DF(g)h = \mathcal{L}(h) := h_t(x,t) - a(x,t)h_{xx}(x,t) + b(x,t)h_x(x,t) + c(x,t)h(x,t)\tag{1}$$ for $h \in X.$ where $a$, $b$ and $c$ depend on $g$ and are uniformly bounded in $Y$ provided the $g$'s are uniformly bounded in $X$ norm.

We also have the bound $$\lVert h \rVert_{C^{2, \alpha}} \leq C\lVert \mathcal{L}(h)\rVert_{C^{0, \alpha}} + C\lVert h(\cdot, 0) \rVert_{C^{2, \alpha}}\tag{2}$$ where the constant $C$ doesn't depend on $T$ and depends only on norms of $g.$

I am told that this bound (2) is

a bound on the mapping norms of the inverses of the Frechet derivatives that doesn't depend on $T$

How does the bound (2) mean that the $DF[g]^{-1}$ are uniformly bounded? Applying the inverse of the Frechet derivative to (1) and using the bound gives $$\lVert DF[g]^{-1}\mathcal{L}(h) \rVert \leq C\lVert \mathcal{L}(h)\rVert_{C^{0, \alpha}} + C\lVert h(\cdot, 0) \rVert_{C^{2, \alpha}}$$ but not sure how to extract the operator norm from here. Can someone help me please?

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