Is it an equivalent norm? I have $$N(f)=\sqrt{f^2(0)+\int_0^1[f'(t)]^2 dt}$$
Iwant to prove that $N$ is a norm on $E=C^1([0,1],\mathbb{R})$
How to do with the third condition:  
$N(f+g)^2= N(f)^2+N(g)^2+2\int_0^1 f'(t)g'(t)dt $
How to find that $N(f+g)\leq N(f)+N(g)$ ?
*** And how to prove that $N$ is equivalent to $||.||_{\infty}$ ?
Thank you 
 A: Note that $N$ is the square root of the sum of the squares of two functions, which are $f\mapsto f(0)$ and $f\mapsto\sqrt{\int_0^1\bigl(f'(t)\bigr)^2\,dt}$. Each one of them is a seminorm and it is easy to deduce from this that $N$ itself is a seminorm. It is in fact a norm, because if $N(f)=0$, then $f(0)=0$ and $f'\equiv0$; therefore, $f$ is the null function.
Here is a direct approach. You want to prove that $N(f+g)\leqslant N(f)+N(g)$, which is equivalent to $N(f+g)^2\leqslant N(f)^2+N(g)^2+2N(f)N(g)$. In other words, you want to prove that$$\begin{multline*}(f+g)^2(0)+\int_0^1\bigl((f+g)'(t)\bigr)^2\,dt\leqslant\\ \leqslant f(0)^2+\int_0^1\bigl(f'(t)\bigr)^2\,dt+g(0)^2+\int_0^1\bigl(g'(t)\bigr)^2\,dt+\\+2\sqrt{f^2(0)+\int_0^1\bigl(f'(t)\bigr)^2\,dt}\sqrt{g^2(0)+\int_0^1\bigl(g'(t)\bigr)^2\,dt}.\end{multline*}$$Expanding the squares and simplifying, this amounts to$$f(0)g(0)+\int_0^1f'(t)g'(t)\,dt\leqslant\sqrt{f^2(0)+\int_0^1\bigl(f'(t)\bigr)^2\,dt}\sqrt{g^2(0)+\int_0^1\bigl(g'(t)\bigr)^2\,dt}.$$But, from Cauchy-Schwarz:$$\int_0^1f'(t)g'(t)\,dt\leqslant\sqrt{\int_0^1\bigl(f'(t)\bigr)^2\,dt}\sqrt{\int_0^1\bigl(g'(t)\bigr)^2\,dt}.$$So, all you need to prove is that$$\begin{multline*}f(0)g(0)+\sqrt{\int_0^1\bigl(f'(t)\bigr)^2\,dt}\sqrt{\int_0^1\bigl(g'(t)\bigr)^2\,dt}\leqslant\\ \leqslant\sqrt{f^2(0)+\int_0^1\bigl(f'(t)\bigr)^2\,dt}\sqrt{g^2(0)+\int_0^1\bigl(g'(t)\bigr)^2\,dt}.\end{multline*}$$This follows again from Cauchy-Schwarz, this time under the form:$$(\forall a,b,c,d\in\mathbb{R}):ac+bd\leqslant\sqrt{a^2+b^2}\sqrt{c^2+d^2}.$$
However, $N$ is not equivalent to $\|.\|_\infty$. In order to see why, take $f_n\colon[0,1]\longrightarrow\mathbb R$ as the function defined by $f_n(t)=t^n$ ($n\in\mathbb N$). Then $(\forall n\in\mathbb{N}):\|f_n\|_\infty=1$, but\begin{align*}N(f_n)&=\sqrt{{f_n}^2(0)+\int_0^1\bigl({f_n}'(t)\bigr)^2\,dt}\\ &=\sqrt{\int_0^1n^2t^{2n-2}\,dt}\\&=\frac n{\sqrt{2n-1}}\to\infty\end{align*}
