Equivalence of p-norms in [itex]C^{0}_{p}[a,b][/itex]

[ELESSAR TELKONT]

Homework Statement

Let $1\leq s<r<\infty$. For what pairs $s,r$ the norms $\left\|\cdot\right\|_{s},\left\|\cdot\right\|_{r}$ are equivalent?

Homework Equations

I have already proven that $\left\|\cdot\right\|_{s}\leq (b-a)^{\frac{r-s}{rs}}\left\|\cdot\right\|_{r}$ and that $\left\|\cdot\right\|_{s}\leq (b-a)^{\frac{1}{s}}\left\|\cdot\right\|_{\infty}$. Of course I have proven, exposing that
$x_{k}(t)=\begin{cases} 1-kt &0\leq t\leq\frac{1}{k}\\ 0 &\frac{1}{k}\leq t\leq 1 \end{cases}$
has a finite infinite-norm for all k, but the 1-norm goes to 0 as k goes infinity, then I can say that there's no c>0 that $\left\|\cdot\right\|_{\infty}\leq c\left\|\cdot\right\|_{s}$, and of course this is true for $[a,b]$ since I can map the [0,1] to the [a,b].

The Attempt at a Solution

I can't imagine some continuous function sequence that may be used as a counterexample to say that I can't bound the s-norms with the r-norms from below. I have done the work as I have proven for the norms in lp sequence spaces and I get that they are equivalent but I suspect they aren't. Please help.

 

[mighty2000]

Thank you for your post. This is a very interesting question and it seems like you have made some good progress in your attempt to solve it. I would like to offer some suggestions and insights that may help you complete your proof.

Firstly, you have correctly proven that for s,r>1, the norms \left\|\cdot\right\|_{s} and \left\|\cdot\right\|_{r} are equivalent. However, this does not cover the case when s or r is less than 1. In order to address this, you should consider the case when s,r\leq 1. You can use the same approach as you did for s,r>1, but you may need to make some adjustments to account for the different ranges of s and r. I suggest starting with the case when s=r=1 and then generalize to s,r\leq 1.

Secondly, you have shown that \left\|\cdot\right\|_{s}\leq (b-a)^{\frac{1}{s}}\left\|\cdot\right\|_{\infty} for all s>1. However, this does not necessarily mean that there is no c>0 such that \left\|\cdot\right\|_{\infty}\leq c\left\|\cdot\right\|_{s}. In fact, for s>1, such a constant does exist. This can be seen by considering the function f(t)=t^{s-1}, which has a finite \infty-norm but an infinite s-norm. However, for s\leq 1, such a constant does not exist and this is an important observation for your proof.

Lastly, for the case when s,r\leq 1, you can use the same function sequence x_{k}(t) that you have defined for s,r>1. However, for s,r\leq 1, you may need to modify the function to ensure that it has a finite s-norm. For example, you can consider the function x_{k}(t)=\frac{1}{k^{\frac{1}{s}}}(1-kt)^{\frac{1}{s}} for 0\leq t\leq\frac{1}{k} and 0 for \frac{1}{k}\leq t\leq