Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.