Composition between a norm and concave function

Let $g:\mathbb{R_+}\rightarrow \mathbb{R}$, defined as follows : $g(x)=\frac{x}{1+x}$, it's concave ( since $g''(x)\leq0$).

Now we define the sup-norm on $C^{\infty}[0,1]$, $\psi(f)=\sup_{x\in[0,1]}|f(x)|$.

I want to prove that $go \psi$ is concave.

But obviously $\psi$ is convex, and the composition of convex and concave function is not necessary concave.

I don 't think this is true. Let $f=1$ and $g=-1$. If $g\circ \psi$ is concave then $g(\psi (\frac {1+(-1)}2) \geq \frac {g(\psi(1)+g(\psi(-1)} 2$. The left side is $0$. but $g(\psi (1))=g(\psi (-1))=\frac 1 2 $ so the right side is $1$.