Strengthening bound in the Euclidean space

Question

Suppose that $z$ is a $K\times 1$ vector. Denote the components of $z$ as $z_1,\ldots, z_K$. Let $r>0$ be given. I'd like to find the smallest constant $C$ such that $$ |z|^r\leq C\cdot\sum_{k=1}^K|z_k|^r\tag{i} $$ holds for all $z$. Here, $|\cdot|$ denotes the absolute value for scalars and the usual Euclidean norm for vectors. $C$ may depend on $r$ and $K$ but not on $z$.

What I have so far: $$ |z|^r=(\sum_k|z_k|^2)^{r/2}\leq(K \max_k|z_k|^2)^{r/2}=K^{r/2}\max_k|z_k|^r\leq K^{r/2}\sum_k|z_k|^r. $$ So $C=K^{r/2}$ works. But can this be improved? (I know that $C$ cannot go strictly below $1$.)

Answer 1

Let's write $\|z\|_r = \left(\sum_{k=1}^K |z_k|^r \right)^{1/r}$; this is called the $r$-norm of $z$. In particular $|z|=\|z\|_2$. We are now in the situation of Relations between p norms. Two cases:

• If $r>2$, then Hölder's inequality gives $\|z\|_2\le K^{1/2-1/r}\|z\|_r$, as shown in the aforementioned Q&A. This is attained by $z=(1,1,1,\dots,1)$.
• If $0<r<2$, then $\|z\|_2\le \|z\|_r$, attained by $(1,0,0,\dots,0)$. The inequality is shown here, but here is a proof again: rescale $z$ so that $\|z\|_r=1$. Since $|z_k|\le \|z\|_r\le 1$, we have $|z_k|^2\le |z_k|^r$, therefore $\sum |z_k|^2\le \sum |z_k|^r= 1$, which means $\|z\|_2\le 1$.

Conclusion: $C=\max(K^{1/2-1/r},1)$.