Energy estimate of grid function

Question

Consider a grid function $v$ defined on $$ \Omega_h:=\{x_i:0\leq i\leq m\} $$ where $x_i=a+ih$, $h=\frac{b-a}{m}$.There are three types of norms $$ ||v||=\sqrt{h\sum_{i=1}^{m-1}v_i^2} $$ $$ ||v||_{\infty}=\mathop{max}\limits_{0 \leq i\leq m}|v_i| $$ $$ |v|_1=\sqrt{h\sum_{i=1}^{m}(\delta_xv_{i-\frac{1}{2}})^2} $$ where $\delta_xv_{i-\frac{1}{2}}=\frac{v_i-v_{i-1}}{h}$.
I am trying to prove that if $v_0=0$ and $v_m$ takes arbitrary values, then $$ ||v||_{\infty}\leq\sqrt{b-a}|v|_1,\quad||v||\leq\frac{b-a}{\sqrt{2}}|v|_1. $$ And if $v_0$ and $v_m$ take arbitrary values then $$ (\max_{0\leq i\leq m}|u_i|)^2\leq2\max\{u_0^2,u_m^2\}+\frac{b-a}{2}|u|_1^2 $$ I have known the case when $v_0=v_m=0$: $$ v_i=\sum_{j=1}^i(v_j-v_{j-1})=h\sum_{j=1}^i\delta_xv_{j-\frac{1}{2}} $$ $$ v_i=-\sum_{j=i+1}^m(v_j-v_{j-1})=-h\sum_{j=i+1}^m\delta_xv_{j-\frac{1}{2}}. $$ Then applying the Cauchy-Schwarz inequality gives $$ |v_i|^2\leq(x_i-a)[h\sum_{j=1}^i\delta_xv_{j-\frac{1}{2}}]^2 $$ $$ |v_i|^2\leq(b-x_i)[h\sum_{j=i+1}^m\delta_xv_{j-\frac{1}{2}}]^2. $$ Then $$ (b-a)|v_i|^2\leq(b-x_i)(x_i-a)|v|_1^2\leq\frac{(b-a)^2}{4}|v|_1^2. $$ The rest of the proof is simple. But I can't figure it out in the case when $v_0$ and $v_m$ take arbitrary values rather than 0.
Can anyone help me?