calculus - Consider the increasing, concave function $x^{0.5}$ on $[0, 1]$.

Consider the increasing, concave function:
$$g(x) = \sqrt x, x ∈ [0, 1].$$

Can you state a continuous function:
$$f(x), x ∈ [0, 1]$$

such that $f(0) = 0, f(x)$ is twice continuously differentiable on $(0, 1]$ and:

$$0 < f'< g', f'' > |g''|$$

for all $x ∈ (0,1]$ ?

So basically I want an increasing function $f(x)$ which has a lower slope than $g(x)$ everywhere but is more convex than $g(x)$ is concave everywhere.