ACTA issues

## When dim$=$const implies subspace$=$const

J. M. Szucs

Acta Sci. Math. (Szeged) 66:3-4(2000), 541-552
2750/2009

 Abstract. If the dimension of the space spanned by the vectors $\langle f_{1}^{(s)}(x)$, $\ldots$, $f_{n}^{(s)}(x)\rangle$, $s=0,1,\ldots,k$, of $n$ real-valued functions $f_{1},\ldots,f_{n}$ and of their first $k$ derivatives is independent of $x\in I$ (an interval $\subseteq{\msbm R}{}$) and is at most $k$, then the space itself is independent of $x\in I$. This was proved by Curtiss and Moszner assuming the continuity of $f_{1}^{(k)},\ldots,f_{n}^{(k)}$. Their proofs are simplified and extended to operator-valued maps. The extension relies on this generalization of a theorem of Peano: Let $T\colon I\to L(V,W)$ be a differentiable map from a nondegenerate interval $I\subseteq{\msbm R}$ to the space $L(V,W)$ of linear operators from a real finite-dimensional vector space $V$ to another such space $W$. Then $\mathop{\rm range}T(x)$, $x\in I$, is constant if and only if $\mathop{\rm range}T^{\prime }(x)\subseteq\mathop{\rm range}T(x)$, $\dim\mathop{\rm range}T(x)=\mathop{\rm const}$, $x\in I$. AMS Subject Classification (1991): 15A15, 26A06, 15A04 Received June 23, 1999, and in revised form December 2, 1999. (Registered under 2750/2009.)