There is the following very general but somewhat tricky theorem—let me phrase it in terms of concave functions because that's the way I'm used to doing it. What follows is a slightly expanded version of an argument given on p. 13f at the beginning of Chapter 3 of Robert R. Phelps's Lectures on Choquet's Theorem, Springer Lecture Notes in mathematics 1757 (2001):
Recall that a function is upper semi-continuous if $\{f \lt c\}$ is open for all $c$.
Let $K$ be a closed convex set in a locally convex space $X$ (you may take $X = \mathbb{R}^n$ if you like, of course). Then a bounded function $f:K \to \mathbb{R}$ is upper semi-continuous and concave if and only if
$$f(x) = \inf{\{a(x)\,:\,a:K \to X \text{ is continuous, affine and } f \leq a\}}$$
for all $x \in K$.
The idea is to use the separating hyperplane theorem. If a function is convex and upper semicontinuous then its subgraph is convex (by concavity) and closed (by upper semicontinuity) and thus the infimum $\hat{f}\,(x)$ over all values $a(x)$ of affine functions dominating $f$ can't lie strictly above the subgraph, for else we could find a separating hyperplane (= the graph of an affine function) lying strictly between the point $(x,\hat{f}(x))$ and the subgraph.
Since a $C^1$ function is continuous, hence upper semi-continuous, the question you ask follows from this immediately. Note also that continuity of affine functions is not an issue if $X = \mathbb{R}^n$.
Here are some more details:
For any bounded function $f$ put $\hat{f}(x) = \inf{\{a(x)\,:\,a:K \to X \text{ is continuous, affine and } f \leq a\}}$. The function $\hat{f}$ is called the concave envelope of $f$. As an infimum of continuous functions, $\hat{f}$ is certainly upper semi-continuous, and as an infimum of concave functions $\hat{f}$ is concave, so the conditions on $f$ are certainly necessary.
Now suppose $f$ is bounded, upper semi-continuous and concave. Consider the space $X \times \mathbb{R}$. Since $f$ is upper semi-continuous and concave, the subgraph $G_f = \{(x,t) \in K \times \mathbb{R}\,:\,f(x) \geq t\} \subset X \times \mathbb{R}$ is closed and convex.
Suppose towards a contradiction that there is a $k \in K$ such that $f(k) \lt \hat{f}(k)$. Then the Hahn-Banach separation theorem (or the separating hyperplane theorem if $X = \mathbb{R}^n$, see also this related thread) gives us a linear functional $\phi: X \times \mathbb{R} \to \mathbb{R}$ and $t_0 \in \mathbb{R}$ such that $$\sup\limits_{(x,t) \in G_f}{\phi(x,t)} \lt t_0 \lt \phi(k, \hat{f}(k)).$$ But $\phi(k,f(k)) \lt \phi(k,\hat{f}(k))$ gives us by linearity of $\phi$ that $\phi(0,\hat{f}(k)-f(k)) \gt 0$ and hence $\phi(0,1) \gt 0$. But this means that for each $x \in X$ there is a unique $a(x) \in \mathbb{R}$ such that $\phi(x,a(x)) = t_0$. It is not hard to show that $a$ is continuous and affine. Since for $x \in K$ we have $\phi(x,f(x)) \lt t_0$ and since $\phi(0,1) \gt 0$ we conclude from the definition of $a$ that for $x \in K$ we have $f(x) \lt a(x)$. On the other hand $t_0 \lt \phi(k, \hat{f}(k))$ but this gives us $f(k) \lt a(k) \lt \hat{f}(k)$, a contradiction to the definition of the concave envelope $\hat{f}$.
Let $(g_i)_{i\in I}$ be a family of convex functions on a convex compact set $\Omega\subseteq \mathbb{R}^d$. We will show that the sup of this family is convex. We will use the standard definition of convexity.
Let $g:=\sup_{i\in I} g_i$.
Take $x,y\in\Omega$ and $t\in[0,1]$.
Fix $i\in I$.
Since $g_i$ is convex and bounded above by $g$, we have
$$
g_i(tx+(1-t)y)\leq tg_i(x)+(1-t)g_i(y)\leq tg(x)+(1-t)g(y).
$$
Since the latter holds for every $i\in I$, we can take the sup and find
$$
g(tx+(1-t)y)\leq tg(x)+(1-t)g(y).
$$
This holds for every $x,y\in \Omega$ and every $t\in[0,1]$.
So $g$ is convex.
Now every affine function $f_i$ is convex, so the result follows from the general case above.
Geometrically? A function is convex iff its epigraph is convex.
See here for a definition of the epigraph.
It is clear that the epigraph of $\sup g_i$ is the intersection of the epigraphs of all the $g_i$.
Now the intersection of convex sets is convex, which yields a more geometric proof of the statement above.
Best Answer
In convex geometry, a supporting line or supporting hyperplane of a convex set $C$ is an affine co-dimension-$1$ subspace $\{y:a^Ty=b\}$ outside of $C$ that touches $C$ in some point or larger facet, i.e., so that $a^TC\le b$ with equality assumed at least once.
From there it is just a twist of thought to compute the constant as $b=\sup a^TC$ and formalize it by giving it a name as a function
$$S_C(x)=\sup\{x^Ty:y\in C\},$$
i.e., $S_C(x)$ is how much you have to shift the plane $\{y:x^Ty=0\}$ to be a supporting hyperplane, thus support function.
Then $C⊂\{y:x^Ty≤S_C(x)\}$. Starting from any even non-convex $C$, the intersection of these supporting half-spaces is the closure of the convex hull of $C$.
As to $S_C$ itself, convex functions are closed under suprema, and since the family of linear functions $x\mapsto x^Ty$, $y\in C$ is a family of convex functions, their pointwise supremum is again convex.