Concave production function implies convex cost function Let's assume we have an increasing production function $f:\mathbb{R^+} \to \mathbb{R^+}$

Now, assume this production function is concave and that the price of input z is fixed (this is a single-input and single output case). I want to show this implies the corresponding cost function $C^f(w,q)$ is convex. 

My thoughts: 

let $z,z' \in \mathbb{R^+}$ where W.L.O.G $z'&gt;z$ and let $f(z)=q$ and $f(z')=q$

Since $f$ is concave, we take $\alpha \in [0,1]$  $s.t$:

$$f(\alpha z +(1- \alpha)z') \geq \alpha f(z) +(1- \alpha)f(z')$$ 

let $\alpha z +(1- \alpha)z'=z''\in \mathbb{R^+}$ where by necessity $z\leq z'' \leq z'$ 

Since cost of $z=w$ is fixed at some $w \in \mathbb{R^+}$

we can rewrite our cost function as $C^f(q)$ which is convex if for $\alpha \in [0,1]$:

$$C(\alpha q +(1- \alpha)q') \leq \alpha c(q) +(1- \alpha)c(q')$$

Now, since $f$ is concave, it cannot be the case that $f$ experiences increasing returns to scale. Then $f$ has constant returns to scale or diminishing returns to scale. 

Then:

Since $z''$ is clearly feasible then we have that:

$$C(w,q'') \leq w*z''$$ 
$$=\alpha w*z + (1- \alpha) w*z'$$
$$= \alpha c(w,q) + (1- \alpha) c(w,q')$$ 

Does this look correct?
Given the fixed input price $w$, the cost function can be written as
$$
C(q)=f^{-1}(q)\times w
$$
where $f^{-1}$ is the inverse of the production function $f$. From the discussion here, one can conclude that the inverse of a concave strictly increasing function is convex. Thus, $C(q)$ is convex as well. 

Going back to your approach, you might like to have this clearly stated. Let $q''=\alpha q + (1-\alpha)q'$, $f(z)=q$, $f(z')=q'$, and $f(z'')=q''$ Then 
$$
\begin{align}
f(z'')&amp;=&amp;q''\\
&amp;=&amp;(\alpha q + (1-\alpha)q')\\
&amp;=&amp;\alpha f(z)+(1-\alpha) f(z')\\
&amp;\leq&amp;f(\alpha z +(1-\alpha z'))
\end{align}
$$
Then $f(z'')\leq f(\alpha z +(1-\alpha z'))$ implying that $\alpha z +(1-\alpha z')\geq z''$ since $f$ is (strictly) increasing. Hence, 
$$
z''w=C(q'')\leq \alpha C(q)+ (1-\alpha) C(q')
$$