Triangle Inequality for norm integral $\|f\|_1=(\int_a^b [|f|^2+|f'|^2]\mathsf dx)^{1/2}$. Define $C_1^1[a,b]$ to be the space of continuously differentiable functions on $[a,b]$, with norm $$\|f\|_1=\left(\int_a^b[|f|^2+|f'|^2]\mathsf dx\right)^{1/2}.$$ Show that this is a proper definition of a norm. Is this normed space complete?
I'm stuck on the triangle inequality part. I'm having a hard time showing that $$\|f+g\|_1\leq \|f\|_1+\|g\|_1.$$ Any hints or solutions are greatly appreciated.
 A: We will assume that $f$ and $g$ are real-valued.  The proof for complex-valued $f$ and $g$ follows analogously.
Let $||f||_1^2\equiv \int_a^b|f|^2dx+\int_a^b|f'|^2dx$.  Then, we have
$$\begin{align}
||f+g||_1^2&=\int_a^b|f+g|^2\,dx+\int_a^b|f'+g'|^2\,dx\\\\
&=\int_a^b(|f|^2+|f'|^2)\,dx+\int_a^b (|g|^2+|g'|^2)\,dx \\\\
&+2\int_a^b(fg+f'g')\,dx \tag 1\\\\
&\le \int_a^b(|f|^2+|f'|^2)\,dx+\int_a^b (|g|^2+|g'|^2)\,dx\\\\
&+2\sqrt{\int_a^b(|f|^2+|f'|^2)\,dx}\sqrt{\int_a^b(|g|^2+|g'|^2)dx} \tag 2\\\\
&=\left(\sqrt{\int_a^b(|f|^2+|f'|^2)\,dx}+\sqrt{\int_a^b(|g|^2+|g'|^2)\,dx}\right)^2\\\\
&=\left(||f||_1+||g||_1\right)^2
\end{align}$$
as was to be shown!  

NOTE:
In going from $(1)$ to $(2)$, we used the Cauchy-Schwarz Inequality.  To prove the inequality explicitly here, we first note that 
$$\int_a^b\left(f-\frac{\int_a^b(fg+f'g')\,dx'}{\int_a^b(g^2+g'^2)\,dx'}\,g\right)^2\,dx+\int_a^b\left(f'-\frac{\int_a^b(fg+f'g')\,dx'}{\int_a^b(g^2+g'^2)\,dx'}\,g'\right)^2 \ge 0$$
since the integrands of both integrals are non-negative.  Then, completing the squares of both integrands and gathering terms yields
$$\left(\int_a^b(f^2+f'^2)\,dx\right)\left(\int_a^b(g^2+g'^2)\,dx\right)\ge \left(\int_a^b(fg+f'g')\,dx\right)^2 \tag 3$$
Taking the square root of both sides of $(3)$, we obtain the desired result
$$\int_a^b(fg+f'g')\,dx\le \sqrt{\left(\int_a^b(f^2+f'^2)\,dx\right)}\sqrt{\left(\int_a^b(g^2+g'^2)\,dx\right)}$$
A: Hint: prove that this norm is induced by an inner product, it will be easier: $$\langle f,g\rangle = \int_a^b f(x)g(x) + f'(x)g'(x)\,{\rm d}x$$
Throw in some conjugates for the complex case. This is the only case that your norm comes from an inner product. In general, we have that $$\|f\|_p = \left( \int_a^b \sum_{j = 0}^k |f^{(j)}(x)|^p \,{\rm d}x \right)^{1/p}, $$ for $1\leq p < \infty$ is a norm in the space of the $C^k$ functions, but this one is harder to prove, you'll have to use some variant of Hölder's inequality, and such.
These spaces are not Banach spaces, as pointed in the comments. 
