How to show that $\|\cdot\|_1$ forms a norm on $\mathcal C^1[0,1]$ I have doubts from the following problem:

Let $\mathcal C^1[0,1]$ denote the set of all continuous real valued functions on $[0,1]$ which are continuously differentiable on $(0,1)$ and whose derivatives can be continuously extended to $[0,1].$ Define $$\|f\|_1=\sqrt{\int_0^1|f(t)|^2\,dt+\int_0^1|f'(t)|^2\,dt}.$$ Show that $\|\cdot\|_1$ defines a norm on $\mathcal C^1[0,1].$
If we define $$|f|_1=\sqrt{\int_0^1|f'(t)|^2\,dt}$$
does $|\cdot|_1$ define a norm on $\mathcal C^1[0,1]?$


*

*I have solved the second part: $|\underline{1}|_1=0$ but $\underline{1}\ne\underline{0}.$ So $|\cdot|_1$ does not form a norm on $\mathcal C^1[0,1].$

But I am clueless about the first part. I tried using Minkowski's inequality to show $$\sqrt{\int_0^1f^2+\int_0^1f'^2}+\sqrt{\int_0^1g^2+\int_0^1g'^2}\ge\sqrt{\int_0^1(f+g)^2+\int_0^1(f'+g')^2}$$
but it didn't work.

How to solve first part?

 A: Maybe you can prove the following:
$$\langle f,g\rangle_1:=\int_0^1 f(x)g(x)\, dx + \int_0^1f'(x) g'(x)\, dx$$
defines a scalar product on $\mathcal{C}^1((0,1))$, then your norm would be induced by $\langle \cdot,\cdot\rangle_1$.
A: Let $p(f)=\|f\|_1=\sqrt{\int_0^1f^2(t)\,\mathrm dt}$ and let $q(f)=\|f'\|_1$. You want to prove that$$(\forall f,g\in\mathcal{C}^1\bigl([0,1]\bigr):\sqrt{p^2(f+g)+q^2(f+g)}\leqslant\sqrt{p^2(f)+q^2(f)}+\sqrt{p^2(g)+q^2(g)}.$$ You know that $p(f+g)\leqslant p(f)+p(g)$ and that $q/f+g)\leqslant q(f)+q(g)$. Therefore$$\sqrt{p^2(f+g)+q^2(f+g)}\leqslant\sqrt{\bigl(p(f)+p(g)\bigr)^2+\bigl(q(f)+q(g)\bigr)^2}$$and threfore it will be enough to prove that$$\sqrt{\bigl(p(f)+p(g)\bigr)^2+\bigl(q(f)+q(g)\bigr)^2}\leqslant\sqrt{p^2(f)+q^2(f)}+\sqrt{p^2(g)+q^2(g)}.$$Squaring both sides and simplifying, this becomes equivalent to$$p(v)p(w)+q(v)q(w)\leqslant\sqrt{p^2(v)+q^2(v)}\sqrt{p^2(w)+q^2(w)}.$$Now, apply the Cauchy-Schwarz inequality and you're done.
A: Notice that $\|f\| = \sqrt{\|f\|_2^2  + \|f'\|_2^2}$ and use the fact that $\|\cdot\|_2$ is a norm so it satisfies the triangle inequality, and CSB for the standard inner product on $\mathbb{R}^2$.
\begin{align}\int_0^1(f+g)^2+\int_0^1\left(f'+g'\right)^2 &= \|f+g\|_2^2 + \left\|f'+g'\right\|_2^2\\
&\le \left(\|f\|_2 + \|g\|_2\right)^2 + \left(\left\|f'\right\|_2 + \left\|g'\right\|_2\right)^2 \\
&= \|f\|^2_2 + 2\|f\|_2\|g\|_2+\|g\|^2_2 + \left\|f'\right\|^2_2 + 2\left\|f'\right\|_2\left\|g'\right\|_2 + \left\|g'\right\|^2_2 \\
&= \|f\|^2_2 + \|f'\|^2_2 + \|g\|^2_2  + \|g'\|^2_2 + 2 \left\langle \left(\left\|f\right\|_2 , \left\|f'\right\|_2\right),\left(\left\|g\right\|_2 , \left\|g'\right\|_2\right) \right\rangle_{\mathbb{R}^2}\\
&\le \|f\|^2 + \|g\|^2 + 2\sqrt{\left\|f\right\|^2_2 + \left\|f'\right\|^2_2}\cdot\sqrt{\left\|g\right\|^2_2 + \left\|g'\right\|^2_2}\\
&=\|f\|^2 + 2\|f\|\|g\| + \|g\|^2\\
&= \left(\|f\| + \|g\|\right)^2\\
&= \left(\sqrt{\int_0^1f^2+\int_0^1f'^2}+\sqrt{\int_0^1g^2+\int_0^1g'^2}\right)^2
\end{align}
Taking the square root of both sides yields the desired inequality.
