Q:

Are these norms on $C^{1}([0,1])$ equivalent?

On the space $C^1[0,1]$ of real valued continuously differentiable functions defined in $[0,1]$, consider the following norms :
$\displaystyle ||f||_{a}=\sup_{0\le x\le 1}f'(x) + \int_0^1|f(x)|\,dx $
and 
$\displaystyle ||f||_{b}=\sup_{0\le x\le 1}f'(x) + |f(0)|\ $
Are these norms equivalent?
I tried to prove this by definition of equivalent norms or contra example,but I failed.

A:

Yes. If $\sup_{x \in [0,1]} |f'(x)| = 0$, $f$ is constant, and the two norms are the same, so we can restrict to $\sup_{x \in [0,1]} |f'(x)| \not = 0$, and then, by multiplicatively scaling, to $\sup_{x \in [0,1]} |f'(x)| = 1$. Then, $$||f||_a = \sup_{x \in [0,1]} |f'(x)|+\int_0^1 |f(x)|dx \le 1+ \int_0^1 |f(0)|+xdx = |f(0)|+\frac{3}{2} = ||f||_b+\frac{1}{2}.$$ Also, using that $\int_0^1 |f(x)|dx = f(c)$ for some $c \in [0,1]$, we get $$||f||_b = \sup_{x \in [0,1]} |f'(x)| + |f(0)| \le 1+|f(c)|+c \le ||f||_a+1.$$ So since $||f||_a,||f||_b \ge 1$, we see $$\frac{1}{2}||f||_a \le ||f||_b \le 2||f||_a.$$

