How to show $[u]_{\alpha;\Omega} \le (\frac{1}{\lambda} +1)[u]_{\alpha;\Omega}^{(\lambda)}$? $\Omega$ is a bounded smooth domain of $\mathbb R^n$.  Letting
$$
[u]_{\alpha;\Omega} = \sup\limits_{x,y\in \Omega  \\~~ x\ne y}    \frac{|u(x)-u(y)|}{|x-y|^\alpha}
$$
and
$$
[u]_{\alpha;\Omega}^{(\lambda)}= \sup\limits_{~~~~~~~~~x,y\in \Omega \\ 0<|x-y|< \lambda \text{diam}\Omega}  \frac{|u(x)-u(y)|}{|x-y|^\alpha}
$$
where $\alpha,\lambda \in (0,1)$. Then,  how to show $[u]_{\alpha;\Omega} \le (\frac{1}{\lambda} +1)[u]_{\alpha;\Omega}^{(\lambda)}$ ?
PS： I really can't deal it. Thanks for any help.
 A: Assuming that $\Omega$ is convex, consider $x$ and $y$ in $\Omega$. Assume that $|x-y|>\lambda\operatorname*{diam}\Omega$, since otherwise there
is nothing to prove. Take
$$
n=\lfloor|x-y|/(\lambda\operatorname*{diam}\Omega)\rfloor+1.
$$
Then
$$
n-1<|x-y|/(\lambda\operatorname*{diam}\Omega)\leq n.
$$
Subdivide the segment joining $x$ and $y$ into $n$ intervals of length $\delta=|x-y|/n$. Note that
$$
\delta=|x-y|/n\leq\lambda\operatorname*{diam}\Omega.
$$
Since the function $t^{1/\alpha}$ is convex,
$$\bigg(\sum_{i=1}^n c_i\bigg)^{1/\alpha}\le n^{1/\alpha-1} \sum_{i=1}^n c_i^{1/\alpha}.$$
By the triangle inequality
$$|u(x)-u(y)|\le \sum_{i=1}^n |u(x_i)-u(x_{i-1})|,$$
where $x_0=x$, $x_n=y$ and the other $x_i$ are the endpoints of the segment. Since $\delta\le \lambda\text{diam}\Omega$, then
$$|u(x)-u(y)|\le \sum_{i=1}^n |u(x_i)-u(x_{i-1})|\le [u]_{\alpha;\Omega}^{(\lambda)} \sum_{i=1}^n |x_i-x_{i-1}|^\alpha.$$
Now let's raise both sides to the power $1/\alpha$ and let's use the first inequality with $c_i= |x_i-x_{i-1}|^\alpha$. We get
$$|u(x)-u(y)|^{1/\alpha}\le \big([u]_{\alpha;\Omega}^{(\lambda)}\big)^{1/\alpha}n^{1/\alpha-1} \sum_{i=1}^n |x_i-x_{i-1}|
\\=\big([u]_{\alpha;\Omega}^{(\lambda)}\big)^{1/\alpha}n^{1/\alpha-1} |x-y|.$$
Raising to power $\alpha$ we get
$$|u(x)-u(y)|\le [u]_{\alpha;\Omega}^{(\lambda)} n^{1-\alpha} |x-y|^\alpha.$$
To conclude, use the fact that
$$
n\leq|x-y|/(\lambda\operatorname*{diam}\Omega)+1\leq\operatorname*{diam}%
\Omega/(\lambda\operatorname*{diam}\Omega)+1\leq1+1/\lambda.
$$
