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I'm considering the space $W\{n,p\}[0,1]$ of functions with $n-1$ continuous derivatives $f^{(n-1)}$ is absolutely continuous and $f^{(n)}$ is in $L^p[0,1]$. The usual norm is the sum of the $p$-norms of each derivative from $1$ to $n$ and the $p$ norm of the function.

Now just consider the $p$ norm of the function + $p$ norm of the $n^{th}$ derivative, i.e.$(\int |f(x)|^p)^{1/p}$ + $(\int |f^{(n)}|^p)^{1/p}$ I want to show that this is equivalent to the usual norm defined above.

This requires finding positive constants and sandwiching this new norm. In one direction it's obvious since the new norm is less than the usual norm for every $x$, say $\|x\|_2 \leq \|x\|_1$. I'm not sure how to show the other direction.

Arctic Char
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Simon
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    Welcome to MSE. I assume you know $\LaTeX$ from the way you had written the post. I think you were not aware of its existence in the site. You can write the way you write on $\LaTeX$ even here. And here's my token +1. Looking forward to your active participation! –  Feb 12 '12 at 20:55
  • My first guess would be to try to prove an extension Theorem for $W^{k, p} $ in the real line and then use that your two norms are equivalent in $W^{k, p} (\mathbb{R}) $. – Jose27 Apr 29 '19 at 03:17

2 Answers2

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Here is a proof for $p>1$. We want to show the existence of $c>0$ such that $$ \|u \|_{W^{n,p}} \le c ( \|u \|_{L^p} + \|D^nu\|_{L^p}) \quad \forall u\in W^{n,p}(0,1). $$ Assume this is not true. Then for every $k$ there is $u_k$ such that $$ \|u_k\|_{W^{n,p}} > k ( \|u_k\|_{L^p} + \|D^nu_k\|_{L^p}). $$ Clearly, $u_k\ne0$. We can rescale the $u_k$'s to achieve $$ 1 = \|u_k\|_{W^{n,p}} > k ( \|u_k\|_{L^p} + \|D^nu_k\|_{L^p}). $$ So $(u_k)$ is bounded in $W^{n,p}$, moreover $u_k \to 0$ in $L^p$ and $D^nu\to 0$ in $L^p$. We can extract a weakly (weak-star for $p=\infty$) converging subsequence (still denoted by same index): $u_k \rightharpoonup^* u$ in $W^{n,p}$. Here, we need $p>1$. By compactness of the embedding, $u_k \to u$ in $W^{n-1,p}$. We already know that the limit is zero: $u=0$, so $\|u_k\|^p_{W^{n-1,p}} \to 0$.

Going back to the initial inequality (for $p<\infty$), proves $$ 1 = \|u_k\|^p_{W^{n,p}} = \|D^nu_k\|_{L^p}^p + \|u_k\|^p_{W^{n-1,p}} \to 0. $$ Similarly, we get for $p=\infty$: $$ 1 = \|u_k\|_{W^{n,\infty}} =\max( \|D^nu_k\|_{L^\infty} , \|u_k\|_{W^{n-1,\infty})} \to 0. $$ In both ways, we get a contradiction.

daw
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By the fundamental theorem of calculus, $f^{(k)}(x) = \int_0^x f^{(n+1)}(y) dy$. Thus $$ \|f^{k}\|_{L^p} =\left [\int_0^1 \left(\int_0^x f^{(k+1)}(y) dy\right)^p\right]^{1/p} $$ Minkowski's integral inequality tells us that $$ \left[\int_0^1 \left(\int_0^x f^{(k+1)}(y) dy\right)^p\right]^{1/p} \leq \int_0^1 \left(\int_0^x |f^{(k+1)}(y)|^{p} dy\right)^{1/p}dx \leq \|f^{(k+1)}\|_{L^p} $$ Now we see by induction that $$ \sum_{k=1}^n \|f^{(k)}\|_{L^p} \leq (n-1)\|f^{(n)}\|_{L^p} $$ So the norm equivalence easily follows. Note that the calculations here are much easier because the total measure space is one, as we're working on $[0,1]$.

Elan B.
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    You've got a typo in the first formula: it should be $f^{(k+1)}$ and not $f^{(n+1)}$. But even with a $k$ the formula doesn't work: you'd need $f^{(k)}(0)=0$. – Hendrik Vogt Feb 13 '12 at 15:52
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    Let me point out that your argument really doesn't work: the inequality you deduce is wrong if $f$ is constant and non-zero. You have to use the norm $|f|_p$ of $f$ itself. (Your argument is possible for $f$ from the Sobolev space $W_0^{n,p}[0,1]$.) – Hendrik Vogt Feb 14 '12 at 14:42