Approximation property for $C^k([0,1]^m)$

Question

This must be well-known, of course, so excuse me my ignorance.

I think, the Banach space $C^k([0,1]^m)$ (of $k$ times smooth functions on a hypercube $[0,1]^m$) must have the approximation property. Could you, please, enlighten me, how (and where) this is proved?

Answer 1

For the case $m=1$ my proof goes as follows. It is well known that $C^k([0,1])\cong\mathbb{R}^k\oplus C([0,1])$ via Taylor series expansion. Since $C[0,1]$ admits a Schauder basis, then so does $C^k[0,1]$. Existence of Schauder basis ensures approximation property. I believe the same route works for the case $m>1$