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 $\mathcal C^1[0,1]?$


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*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?


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 $\langle \cdot,\cdot\rangle_1$.
