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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.

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By the fundamental theorem of calculus, $f^{(k)}(x) = \int_0^x f^{(n+1)}(y) dy$. Thus $$||f^{k}||_{L^p} = [\int_0^1 (\int_0^x f^{(k+1)}(y) dy)^p]^{1/p}$$ Minkowski's integral inequality tells us that $$ [\int_0^1 (\int_0^x f^{(k+1)}(y) dy)^p]^{1/p} \leq \int_0^1 (\int_0^x |f^{(k+1)}(y)|^{p} dy)^{1/p}dx \leq ||f^{(k+1)}||_{L^p}$$ Now we see by induction that $$\sum_1^n ||f^{(k)}||_{L^p} \leq (n-1)||f^{(n)}||_{L^p}$$ So the norm equivalence easily follows. Note that the calculations here are much easier because the total measure space is 1, as we're working on [0,1].

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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
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