# There exist constants such that $\|v\|_{0,2}≤c_1\|v\|_{1,2}+c_2 |u|$

How can I show that there exist constants $c_1,c_2 >0$ such that for all $v \in H^1(a,b)$ the inequality

$$\|v\|_{0,2}≤c_1\|v\|_{1,2}+c_2 |u|$$

holds where $u=(b-a)^{-1}\int_a^bv(\xi)\ d\xi$?

I think I need to use the Hölder inequality, or so.

Some help is much appreciated.

• Can you please define the $0,2$ and $1,2$ norms(?).
– Alt
Commented May 16, 2018 at 17:16
• $||u||_{0,2}=(∫_a^b u(x)^2 dx)^{1/2}$ and $||u||_{1,2}=(∫_a^b u'(x)^2 dx)^{1/2}$ Commented May 16, 2018 at 17:19

For any $x, y \in [a,b]$, and $x < y$ we have $$|v(x) -v(y)| = |\int_x^y v'(s) ds| \leq (y-x)^{\frac12} \left(\int_x^y v'(s)^2 ds\right)^{\frac12} \leq (y-x)^{\frac12} \left(\int_a^b v'(s)^2 ds\right)^{\frac12}.$$ Hence $v \in C([a,b])$. By mean value theorem, there exists $x_0\in [a,b]$ such that $v(x_0) = (b-a)^{-1} \int_a^b v(s) ds$. For $x \in [x_0,b]$, we have $v(x) = v(x_0) + \int_{x_0}^x v'(s) ds$, hence $$|v(x)| \leq |v(x_0)| + (x-x_0)^{\frac12} (\int_{x_0}^x v'(s)^2 ds)^{\frac12} \leq |v(x_0)| + \|v\|_{1,2} (x-x_0)^{\frac12}.$$ Using the inequality $(s+ t)^2 \leq 2(s^2 + t^2)$ we have $$\int_{x_0}^b v(x)^2 dx \leq 2(b-x_0) v(x_0)^2 + 2 \|v\|_{1,2}^2 \int_{x_0}^b (x-x_0) dx \\= 2(b-x_0) v(x_0)^2 + \|v\|_{1,2}^2 (b-x_0)^2.$$ Similarly, $$\int_a^{x_0} v(x)^2 dx \leq 2(x_0-a) v(x_0)^2 + 2 \|v\|_{1,2}^2 \int_a^{x_0} (x_0-x) dx \\= 2(x_0-a) v(x_0)^2 + \|v\|_{1,2}^2 (x_0-a)^2.$$ Thus, by the inequality $s^2 + t^2 \leq (s+t)^2$ for $s,t\geq 0$, we get $$\int_a^b v(x)^2 dx \leq 2(b-a) v(x_0)^2 + \|v\|_{1,2}^2 (b-a)^2.$$ Using the inequality $\sqrt{s+ t} \leq \sqrt{s} + \sqrt{t}$, we get $$\|v\|_{0,2} \leq (b-a) \|v\|_{1,2} + \sqrt{2(b-a)} |\frac1{b-a} \int_a^b v(x) dx|.$$
• Thanks for the help! but is it "by mean value theorem .... $v(x_0)=(b-a)^{-1} ∫_a^b v(s)ds=u(s)$? Commented May 17, 2018 at 9:55