$p$-norm inequality

Question

Let $p \ge 2$ and $q$ such that$${1\over p} + {1\over q} = 1.$$Is it true that there exists a constant $c$ such that for all $x$, $y$ such that $\|x\|_q \le 1$ and $\|y\|_q \ge 1$$$\left\|x - {y\over{\|y\|_q}}\right\|_p \le c \cdot \|x - y\|_p?$$Note that for $p = 2$, it is known that $c = 1$. When $p = \infty$, $c$ can not be $1$ due to the counterexample $x = (1, 0, 0)$, $y = (.5, .5, .5)$.

Answer 1

Let us assume that we are working with sequence spaces $\ell^p$ and $\ell^q$. Since the$$p=q=2$$case is taken care of already, assume$$p>2>q.$$Since $\ell^p\subset\ell^q$ for $q\le p$ as here, both sides of the desired inequality make sense. Start the indices of the sequences at zero, because we feel like it. So, an example to stretch things: $$x_k=\begin{cases}1&k=0\\0&k\ge 1,\end{cases}$$ $$y_k=\begin{cases}1&k=0\\a&1\le k\le n\\0&k>n,\end{cases}$$ where $a\in (0,1)$ and $n\ge 1$ will be chosen later. Then$$\|x-y\|_p=a\sqrt[p]{n},$$$$\|y\|_q=\sqrt[q]{1+na^q},$$and $$\left\|x-\frac{y}{\|y\|_q}\right\|_p=\left(\left(1-\frac1{\|y\|_q}\right)^p+n\left(\frac{a}{\|y\|_q}\right)^p\right)^{\frac1p}=\frac{\left(\left(\|y\|_q-1\right)^p+na^p\right)^{\frac1p}}{\|y\|_q}.$$ Note that the $na^p$ term there is $\|x-y\|_p^p$. That is not going to help in making the whole thing big relative to $\|x-y\|_p$. We need instead to have $\|x-y\|_p$ small while simultaneously $\|y\|_q-1$ is relatively large. To do this, choose $n$ large and$$a=2\sqrt[q]{\frac1n}.$$Then $na^q=2^q$ and $\|y_q\|> 2$. From this,$$\left\|x-\frac{y}{\|y\|_q}\right\|_p>\frac{((\|y\|_q-1)^p)^{\frac1p}}{\|y\|_q}>\frac12.$$ On the other hand,$$\|x-y\|_p=2\cdot\sqrt[q]{\frac1n}\cdot\sqrt[p]{n}=2\cdot n^{\frac1p-\frac1q}.$$Since $q<p$, that exponent is negative. As $n\to\infty$, the whole quantity then goes to zero. So then, is there a constant $c$ such that (something $> 1/2$) is at most ($c$ times something that goes to zero)? No, there is not. The needed $c$ goes to $\infty$ as $n$ does.

The reciprocal nature of $p$ and $q$ was not used, only the fact that $p>q$ was.