3

Consider the $p$-norm with weight factors over $\mathbb R^n$: $$ \| x \|_{p,\lambda} := \Big( \lambda_1 |x_1|^p + \lambda_2 |x_2|^p + \dots + \lambda_n |x_n|^p \Big)^{\frac 1 p} $$ Here, $\lambda_i > 0$ for all $ 1 \leq i \leq n$.

Suppose that $1 \leq p < q$. I am looking for the best constant $C$ in the inequality $$ \| x \|_{q,\lambda} \leq C \| x \|_{p,\lambda} $$ but I struggle with proving that. Here, $C$ may depend on $n$ and $\lambda$.

When $\lambda_1 = \dots = \lambda_n = 1$, then the best constant $C = 1$ holds, and I suspect that $C = 1$ is possible also for general positive weights.

shuhalo
  • 8,084
  • 1
    In the first equation did you miss the $(\cdot)^{1/p}$? Without being raised to the $\frac1p$ it's not a norm in general. – Vim Nov 19 '24 at 15:46
  • Assuming you're talking about the usual $L^p$ norm, this post seems very relevant to your problem. You might start from the Holder's inequality mentioned in the accepted answer there. – Vim Nov 19 '24 at 15:50
  • @Vim thanks, typo – shuhalo Nov 19 '24 at 16:15

1 Answers1

1

The best constant is equal $$C=\max_{1\le k\le n}\lambda_k^{1/q-1/p}$$ Indeed, let $\{e_k\}_{k=1}^n$ denote the standard basis in $\mathbb{R}^n.$ We have $$\|e_k\|_{p,\lambda}=\lambda_k^{1/p},\, \ \|e_k\|_{q,\lambda}=\lambda_k^{1/q}$$ Thus $$C\ge \max_k{\|e_k\|_{q,\lambda}\over \|e_k\|_{p,\lambda}}=\max_{1\le k\le n}\lambda_k^{1/q-1/p}$$ Next, basing on $\|v\|_q\le \|v\|_p$, we get for $v_k=\lambda_k^{1/q}x_k$ $$\|x\|_{q,\lambda}=\left (\sum_{k=1}^n\lambda_k|x_k|^q\right )^{1/q} =\left (\sum_{k=1}^n(\lambda_k^{1/q}|x_k|)^q\right )^{1/q}\\ \le \left (\sum_{k=1}^n(\lambda_k^{1/q}|x_k|)^p\right )^{1/p}=\left (\sum_{k=1}^n\lambda_k^{p/q-1}\lambda_k|x_k|^q\right )^{1/p}\\ \le \max_k\lambda_k^{1/q-1/p}\|x\|_{p,\lambda}$$ Thus $C\le \displaystyle\max_k\lambda_k^{1/q-1/p}.$

Ryszard Szwarc
  • 44,675