Equivalence of Hölder norms 
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...?
 A: 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 
$$ \|f\|_\infty + \left|\frac{f(x+h)-f(x)}{|h|} \right| \le  |f|_1  \le c(\|f\|_\infty + |h|\|f''\|_\infty)~~~~\forall ~~h $$

