prove that space $V$ with norm $\|\varphi\|$ is normed linear space?

Question

I have assignment a norm $$ \|\varphi\|=\left( \int_a^b |\varphi(x)|^2 + |\varphi'(x)|^2 \right)^{1/2} $$ in space $V=C^1(I)$ and I should prove that this space is normed vector space. I know these three rules that $$ \|x\|=0 \Leftrightarrow x=0 \\ \|\alpha x\|=|\alpha| \|x\| \quad ...x \in V,\alpha \in \mathbb{R} \\ \|x+y\| \leq \|x\|+\|y\|$$ but it was told me that the last one (triangle inequality) is not so easy in this case but I can prove it if I use scalar product $ \|\varphi\|=(\varphi,\varphi)^{1/2}$. Is it some way out of this? Because I don't know how..

Answer 1

Define a sequilinear map $\langle\cdot,\cdot\rangle:C^1(I)\times C^1(I)\to\mathbb C$ by $$\langle\varphi,\psi\rangle=\int_a^b\varphi(t)\overline\psi(t)+\varphi'(t)\overline\psi'(t)\ dt$$ Then by showing that $\langle\cdot,\cdot\rangle$ is an inner product on $C^1(I)$, and that $\|\varphi\|=\langle\varphi,\varphi\rangle^{1/2}$, you will have the result (assuming you can show that such a norm satisfies the triangle inequality).

Answer 2

Here is a trick for this sort of stuff (elementary, but noisy):

You know that $\|f\|_{L^2} = \sqrt{\int |f(x)|^2 dx}$ and $\|x\|_{\mathbb{R}^2} = \sqrt{|x_1|^2+|x_2|^2}$ are norms.

Note that both norms are monotone in the sense that if $|f(x)| \le |g(x)|$ then $\|f\|_{L^2} \le \|g\|_{L^2}$ and similarly for the other norm.

Let $T \phi = (\phi, \phi')$, and note that $\|\phi\| = \| x \mapsto \|T \phi(x) \|_{\mathbb{R}^2}\|_{L^2}$, and so \begin{eqnarray} \|\phi_1+\phi_2\| &=& \| x \mapsto \|T (\phi_1+\phi_2)(x) \|_{\mathbb{R}^2}\|_{L^2} \\ &\le & \| x \mapsto \|T \phi_1(x)+T\phi_2(x) \|_{\mathbb{R}^2}\|_{L^2} \\ &\le& \| x \mapsto (\|T \phi_1(x) \|_{\mathbb{R}^2} + \|T \phi_2(x) \|_{\mathbb{R}^2}) \|_{L^2} \\ &\le& \| x \mapsto \|T \phi_1(x) \|_{\mathbb{R}^2}\|_{L^2} + \| x \mapsto \|T \phi_2(x) \|_{\mathbb{R}^2}\|_{L^2} \\ &=& \|\phi_1\|+ \|\phi_2\| \end{eqnarray}