grandes-ecoles 2013 Q5d

grandes-ecoles · France · x-ens-maths2__mp Proof Direct Proof of an Inequality
We denote $\mathcal{I} = \{(j, k) \in \mathbf{N}^{2} \mid j \in \mathbf{N} \text{ and } 0 \leq k < 2^{j}\}$; for $j \in \mathbf{N}$, $\mathcal{T}_{j} = \{k \in \mathbf{N} \mid 0 \leq k < 2^{j}\}$. For all $(j, k) \in \mathcal{I}$, $\theta_{j,k} : [0,1] \rightarrow [0,1]$ is defined by $$\theta_{j,k}(x) = \left\{ \begin{array}{l} 1 - |2^{j+1} x - 2k - 1| \quad \text{if } x \in [k 2^{-j}, (k+1) 2^{-j}] \\ 0 \text{ otherwise} \end{array} \right.$$ Prove that for all $(j, k) \in \mathcal{I}$ and $(x, y) \in [0,1]^{2}$, we have $$|\theta_{j,k}(x) - \theta_{j,k}(y)| \leq 2^{j+1} |x - y|$$
We denote $\mathcal{I} = \{(j, k) \in \mathbf{N}^{2} \mid j \in \mathbf{N} \text{ and } 0 \leq k < 2^{j}\}$; for $j \in \mathbf{N}$, $\mathcal{T}_{j} = \{k \in \mathbf{N} \mid 0 \leq k < 2^{j}\}$. For all $(j, k) \in \mathcal{I}$, $\theta_{j,k} : [0,1] \rightarrow [0,1]$ is defined by
$$\theta_{j,k}(x) = \left\{ \begin{array}{l} 
1 - |2^{j+1} x - 2k - 1| \quad \text{if } x \in [k 2^{-j}, (k+1) 2^{-j}] \\
0 \text{ otherwise}
\end{array} \right.$$
Prove that for all $(j, k) \in \mathcal{I}$ and $(x, y) \in [0,1]^{2}$, we have
$$|\theta_{j,k}(x) - \theta_{j,k}(y)| \leq 2^{j+1} |x - y|$$