Will $\sqrt{h \sum_{i =0}^{N-1} （1 - u_i^2)^2} < C_1$ imply there exist $C_2$ satisfies $\sqrt{h \sum_{i=0}^{N-1} u_i^2} < C_2$

Question

Assume the interval $[a, b]$ is divided by uniform grids $x_i = a + i * h, i = 0, 1, \cdots, N$, where $h = \frac{b - a}{N}$, $\mathbf u = [u_0, \cdots, u_{N-1}]^T$ is a grid funciton, will the boundedness of $\|1 -\mathbf u^2\|_{l^2}$, i.e. \begin{align*} \sqrt{h\sum_{i = 0}^{N-1} (1 - u_i^2)^2} \leq C_1 \end{align*} impliy the boundedness of $\|\mathbf u\|_{l^2}$, i.e. there exists $C_2$ satisfies \begin{align*} \sqrt{h \sum_{i = 0}^{N-1} u_i^2} \leq C_2 \end{align*}

Answer 1

$ a \in \mathbb R^N \mapsto (\sum_{i=1}^N a_i^2)^{1/2}$ is a norm on $\mathbb R^N$. In particular, it satistifies the triangle inequality. Therefore $$ \| u^2\| \leq \| u^2 - 1\| + \| 1\| \leq C_1 + \sqrt {b-a} $$ and by Cauchy-Schwarz inequality $$ \| u\|^2 \leq \| 1 \| \|u^2\| = (b-a)\|u^2\| \leq (b-a)(C_1 + \sqrt {b-a}) $$ so $$ \| u\| \leq C_2 $$ with $C_2 = \sqrt{(b-a)(C_1 + \sqrt {b-a})}$.

Details for the use of Cauchy-Schwarz inequality : $$ \| u\|^2 = h \sum_i u_i^2 \leq h \left(\sum_i 1^2 \right)^{1/2} \left(\sum_i u_i^4 \right)^{1/2} = \| 1\| \| u^2 \| $$