I am trying to solve the following exercise:
Let $W^{1,2}=W^{1,2}(\mathbb R)$ be the Sobolev space consisting of functions $f\in L^2(R)$ such that exists $f_w\in L^2(R)$ with the following condition: $$ \int_R f(x)\, \varphi' (x)\,dx=-\int_R f_w(x)\, \varphi (x)\,dx\qquad \forall \varphi \in \mathcal S(R) $$ where $S(R)$ is the Schwartz space.
There are different things to show but I'm having trouble with the following:
Show that $$\|f\|_ {W^{1,2}}=\left(\|f\|_2^2+\|f_w\|_2^2\right)^{1/2} $$ is a norm defined on su $W^{1,2}$. Obviously $||.||_2 $ is the norm for $L^2-space$ defined as: $\|f\|_ 2 = (\int_R |f(x)|^2dx)^{1/2} $
There are 3 things to prove:
$\|f\|_ {W^{1,2}}=0$ iff $f=0 \Rightarrow 0=\|f\|_ {W^{1,2}}=\left(\|f\|_2^2+\|f_w\|_2^2\right)^{1/2}=0 $ It was easy to prove using that $\|.\|_2$ is a norm.
$\|\lambda f\|_ {W^{1,2}}=|\lambda|\cdot \|f\|_ {W^{1,2}}$ $ \forall \lambda \in R $ $ \Rightarrow \|\lambda f\|_ {W^{1,2}}= \left(\|\lambda f\|_2^2+\|\lambda f_w\|_2^2\right)^{1/2} = \left( |\lambda|^2 \| f\|_2^2+ |\lambda|^2 \| f_w\|_2^2\right)^{1/2} = |\lambda| (\left(\|f\|_2^2+\|f_w\|_2^2\right)^{1/2}) = |\lambda|\cdot \|f\|_ {W^{1,2}} $
$ \|f+h\|_ {W^{1,2}} \le \|f\|_ {W^{1,2}} + \|h\|_ {W^{1,2}}$ This is where I always get stuck: I thought about using once again the norm for $L^2-space$ and\or squaring everything but it seems useless. Can someone help me?
Thanks in advance!
Use that $x \mapsto x^2$ is increasing on the line $[0, \infty)$ and thus $\lVert f + g \rVert_2^2 \leq (\lVert f \rVert_2 + \lVert g \rVert_2)^2$.
Now we know that for some numbers $a_1, b_1, a_2, b_2 \in \mathbb{R}$ we always have: $$ \sqrt{(a_1+b_1)^2+(a_2+b_2)^2} \leq \sqrt{a_1^2 + a_2^2} + \sqrt{b_1^2 + b_2^2} $$ This is Minkowski's inequality, which is basically the triangle inequality that belongs to the euclidian norm.
I am sure that you know how to conclude.