A quantitative version of the converse Taylor theorem: \(C^{k,\omega }\)-smoothness (Q2875417)

Let \(X\) and \(Y\) be normed linear spaces, \(U\subseteq X\) be an open set, and \(f:X\to Y\) be a mapping. Denote \(\mathbb N:=\{1,2,\dots\}\), \(\mathbb N_0=\mathbb N\cup \{0\}\).NEWLINENEWLINEThe following are the main results in the paper.NEWLINENEWLINETheorem 1. Fix also \(k\in \mathbb N_0\). Then \(f\in C^k(X;Y)\) iff \(f\) is a \(T^k\)-smooth mapping with \(\|R(y,h)\|/\|h\|^k \to 0\) as \((y,h)\to (x,0)\), \(h\neq 0\), for every \(x\in U\), where \(R(x,h)=f(x+h)-P^x(h)\), \(P^x=\sum_{j=0}^k(1/j!)d^jf(x)\).NEWLINENEWLINETheorem 2. Fix also \(k\in \mathbb N\). In addition, suppose that \(f\) is \(UT^k\)-smooth on \(U\), and the modulus \(\omega\) from the definition of \(UT^k\)-smoothness is subadditive. Let also \(V\subset U\) be an open bounded subset with \(\text{dist}(V,X\setminus U)> 0\). Then \(f\) is \(C^{k,m\omega}\)-smooth on \(V\), where \(m> 0\) depends on \(k\), \(\text{diam}(V)\) and \(\text{dist}(V,X\setminus U)\). Moreover, if \(U=X\) or \(U\) is a convex bounded set, then (i) \(f\) is \(C^{k,m\omega}\)-smooth on \(U\), (ii) \(\omega\)=subadditive is superfluous, (iii) \(m=c_ke^k_U\), where \(c_k> 0\) is a constant depending on \(k\) only.NEWLINENEWLINEFurther aspects motivated by these statements are also discussed.