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Let $0 <\alpha \leq 1$. Define $$ |f|_1= \sup_{x\in\mathbb{R}}|f(x)|+ \sup_{x,h\in\mathbb{R}}\left|\frac{f(x+h)-f(x)}{|h|^{\alpha}}\right|$$ $$ |f|_2= \sup_{x\in\mathbb{R}}|f(x)|+ \sup_{x,h\in\mathbb{R}}\left|\frac{f(x+h)-2f(x)+f(x-h)}{|h|^{\alpha}}\right| $$ for $f\in C(\mathbb{R}^d)$.

Prove that for every $\alpha\in (0,1)$ and $ f\in C_c^{\infty}(\mathbb{R}^d)$ there exists a constant $c \geq 1 $ s.t. $$ c^{-1}|f|_2\leq |f|_1\leq c|f|_2 $$ Discuss the same question for $\alpha = 1$.

Can anyone help me and give me a hint...?

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  • $\begingroup$ $x \in \mathbb{R}$ in the supremum but $f \in \mathcal{C}(\mathbb{R}^d)$ so $d=1$? $\endgroup$ – Rafael Wagner Oct 23 '17 at 17:54
  • $\begingroup$ oh, I didnt notice that before. I think $ x\in \mathbb{R}^d$ is meant in the supremum. $\endgroup$ – user494831 Oct 23 '17 at 18:00
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First it is easy to see that, $$\|f\|_{\infty} +\left|\frac{f(x+h)-2f(x)+f(x-h)}{|h|^{\alpha}}\right| =\|f\|_\infty+ \left|\frac{f(x+h)-f(x) -(f(x)-f(x-h))}{|h|^{\alpha}}\right|\\\le \|f\|_\infty+\left|\frac{f(x+h)-f(x)}{h^\alpha} \right|+\left|\frac{(f(x)-f(x-h))}{|h|^{\alpha}}\right| \le 2|f|_1 $$

Which implies, by taking the supremum that, $$|f|_2 \le 2|f|_1$$ this means that, if we define the embedding

Next we consider $z= x+k$ and $h=2k$ hence, $x+h =z+k$ and $x =z-k$

Then, for $0<\alpha <1,$ $$\left|\frac{f(x+h)-f(x)}{|h|^{\alpha}}\right| = \left|\frac{f(z+k)-2f(z)+f(z-k) +\color{red}{2 [f(z) -f(z-k)]}}{|2k|^{\alpha}}\right|\\ \le 2^{-\alpha} \left|\frac{f(z+k)-2f(z)+f(z-k)}{|k|^{\alpha}}\right| + 2^{1-\alpha} \left|\frac{\color{red}{ [f(z) -f(z-k)]}}{|k|^{\alpha}}\right|$$

passing through the supremum yields, $$ |f|_1 \le 2^{-\alpha}|f|_2 +2^{1-\alpha} |f|_1 \implies (1-2^{1-\alpha})|f|_1 \le 2^{-\alpha}|f|_2 $$ that is

$$ |f|_1 \le \frac{2^{-\alpha}}{(1-2^{1-\alpha})}|f|_2 $$

Let $$\color{blue}{c=\max(2,\frac{2^{-\alpha}}{(1-2^{1-\alpha})})}. $$ Thus, $$c^{-1}|f|_2\le |f|_1 \le c |f|_2~~~for ~~all ~~f$$

If $\alpha =1$ then for $f\in C^\infty_c(\Bbb R)$ we that $$ f(x+h)-f(x) =\int_0^1 f'(x+th) h dt$$ that is $$ \left|\frac{f(x+h)-f(x)}{|h|} \right| = \left|\int_0^1 f'(x+th) dt.\right|$$

Unlike, for the second case applying the above trick twice we have,$$ \left|\frac{f(x+h)-2f(x)+f(x-h)}{|h|}\right| =\left|\int_0^1 f'(x+th)- f'(x-th)dt.\right|\\ = \left|\int_0^1t\int_0^1 f''(x-th+2sth)h\right| \le |h|\|f''\|_\infty$$

If we assume that $|f|_1 \le c |f|_2 $ then this leads to $$ \|f\|_\infty + \left|\frac{f(x+h)-f(x)}{|h|} \right| \le |f|_1 \le c(\|f\|_\infty + |h|\|f''\|_\infty)~~~~\forall ~~h $$

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    $\begingroup$ you shouldn't prove that $c=2$ as the question of OP? $\endgroup$ – Rafael Wagner Oct 23 '17 at 18:58
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    $\begingroup$ But in the question he says to prove that $c^{-1}|f|_2\le |f|_1 \le c |f|_2$. $\endgroup$ – Rafael Wagner Oct 23 '17 at 19:06
  • $\begingroup$ It is straight forward but I don't see it as irrelevant. Also made your question much more elegant in my opinion. $\endgroup$ – Rafael Wagner Oct 23 '17 at 19:22
  • $\begingroup$ Ok sorry it was not appropriate as word to use $\endgroup$ – Guy Fsone Oct 23 '17 at 19:25
  • $\begingroup$ Thanks for your solution. Two questions: 1) Is there any reason why $f\in C_c^{\infty} $ in the statement? You didnt use it or? 2) Is there a problem for $\alpha =1$? $\endgroup$ – user494831 Oct 24 '17 at 9:49