Holder Space series term

Question

Holder space $C^{k,\gamma}(\bar{U})$ has the norm $||u||_{C^{k,\gamma}(\bar{U})} := \sum_{|\gamma | \leq k}||D^{\alpha}u||_{C(\bar{U})} + \sum_{|\gamma|=k}[D^{\alpha}u]_{C^{0,\gamma}(\bar{U})}$, where this is finite. How would you interpret the second term which is a series, what does this series mean exactly? Thanks

Answer 1

The formula you gave has a typo: $\gamma$ should be replaced by $\alpha$ under the sums. $$||u||_{C^{k,\gamma}(\bar{U})} := \sum_{|\alpha | \leq k}||D^{\alpha}u||_{C(\bar{U})} + \sum_{|\alpha|=k}[D^{\alpha}u]_{C^{0,\gamma}(\bar{U})}$$

Consider functions of one variable first: then the norm consists of suprema of derivatives of orders $\le k$, plus the $C^\gamma$ norm of the $k$th derivative. The last terms is the essential one; the suprema are thrown in so that polynomials won't end up with zero norm. You can use simply $\|f^{(k)}\|_{C^\gamma}$ if you don't mind having a seminorm instead of a norm.

When $f$ has $n$ variables, the notion of "the $k$th derivative" is more complex. Partial derivative are indexed by $n$-tuples $\alpha=(\alpha_1,\dots ,\alpha_n)$ which tell us how many times to take derivative in each variable. The order of the derivative is the sum $\alpha_1+\dots +\alpha_n $. The second summation in the formula runs over all derivatives of order $k$.

For example, with $n=2$ and $k=3$ the sum $\sum_{|\alpha|=k}$ has four terms, corresponding to $\alpha$ being $(3,0)$, $(2,1)$, $(1,2)$, and $(0,3)$.