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I've been reading this answer but I'm stuck on the step where $$\limsup_{p \to \infty} \left( \|x\|_{\infty}^{1-\frac{q}{p}} \cdot \|x\|_q^{\frac{q}{p}}\right) = \|x\|_{\infty} \cdot 1$$ for $1 \le p <\infty, q<p$.

I'm not sure why this holds. Can someone explain this?

Zero Pancakes
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1 Answers1

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For $x$ and $q$ fixed we have $\displaystyle\limsup_{p\to\infty}\|x\|_\infty^{1-\frac{q}{p}}=\lim_{p\to\infty}\|x\|_\infty^{1-\frac{q}{p}}=\|x\|_\infty$

For $x$ and $q$ fixed we also have $\displaystyle\limsup_{p\to\infty}\|x\|_q^\frac{q}{p}=\lim_{p\to\infty}\|x\|_q^\frac{q}{p}=\|x\|_q^0=1$

The previous limits can be obtained like that because $\|x\|_\infty$ and $\|x\|_q$ are just some numbers $a,b\ge0$, so it's the same as thinking of $\displaystyle\lim_{p\to\infty}a^{1-\frac{q}{p}}$ and $\displaystyle\lim_{p\to\infty}b^\frac{q}{p}$.

Since both limits exist, then the limit of the product is the product of the limits, and therefore the $\displaystyle\limsup_{p \to \infty} \left( \|x\|_{\infty}^{1-\frac{q}{p}} \cdot \|x\|_q^{\frac{q}{p}}\right) =\lim_{p\to\infty}\left( \|x\|_{\infty}^{1-\frac{q}{p}} \cdot \|x\|_q^{\frac{q}{p}}\right)= \lim_{p\to\infty} \|x\|_{\infty}^{1-\frac{q}{p}} \cdot\lim_{p\to\infty} \|x\|_q^{\frac{q}{p}}=\|x\|_{\infty} \cdot 1$

Darsen
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  • Why is $\displaystyle\limsup_{p \to \infty} \left( |x|{\infty}^{1-\frac{q}{p}} \cdot |x|_q^{\frac{q}{p}}\right) =\lim{p\to\infty}\left( |x|_{\infty}^{1-\frac{q}{p}} \cdot |x|_q^{\frac{q}{p}}\right)$ ? – Zero Pancakes Oct 25 '20 at 14:45
  • At first $\limsup$ and $\lim$ don't have to be the same, since $\limsup$ is always defined for a bounded sequence, but $\lim$ is not (it could oscillate). But in the case that $\lim$ does exist, i.e., the sequence is convergent, then every subsequence must converge to the same limit. $\limsup_{n\to\infty}x_n$ considers a subsequence of $(x_n){n\in\Bbb N}$; specificly it is the limit of $(\inf{m\ge n}{x_m})_{n\in\Bbb N}$. – Darsen Oct 25 '20 at 16:06