Let $f:\mathbb{R}^n\to \mathbb{R}$ be a convex function. It is well-known that (see, e.g., here) that there exists a set $A$ such that $$ f(x)=\sup_{a\in A}A(a)x+b(a), \quad \forall x\in \mathbb{R}^n, $$ where $A(a)\in\mathbb{R}^{1\times n}$ and $b(a)\in \mathbb{R}^n$.
I was wondering if there is any reference saying we can choose $A$ to be countable?
This is easy. $\{(A(a),b(a)): a \in A\}$ is a subset of a separable metric space, so it is itself separable. Take a countable dense subset of this and use the fact that supremum of any set of real numbers is same as the supremum over countable dense subset.
To complement the other answer, one can prove the existence of $A(a)$ using Hahn Banach.
Indeed, consider $D=\{(x,c)\::\: c>f(x)\}$ and $P=\{(x_o,f(x_o))\}$. By our definition, $P\cap D=\emptyset$.
Because $f$ is convex, it is not difficult to see that $D$ is convex. $D$ is also open, as being convex implies continuity of $f$. Therefore, Hahn Banach (as stated in Brezis functional analysis textbook, but we do not really need openness in $\mathbb{R}^n$) applies and we obtain $(\tilde{A}(a),b)$ such that:
$$\langle (\tilde{A}(a),b),(x_o,f(x_o))\rangle\leq \langle (\tilde{A}(a),b),(x, c)\rangle\quad \forall c>f(x)$$
Rearranging:
$$\langle \tilde{A}(a),x_o-x\rangle+bf(x_o)\leq b c $$
If $b=0$, then $\tilde{A}(a)\not=0$ (Hahn Banach yields a non zero functional). But then, taking $x=x_o+\lambda \tilde{A}(a)$ yields a contradiction by sending $\lambda$ to infinity.
Therefore, $b$ is non zero. If $b<0$ a similar contradiction is found by sending $c$ to infinity. Therefore, $b>0$. Dividing by $b$ and sending $c$ to $f(x)$, one obtains that by defining $A(a)=\tilde{A}(a)/b$:
$$ \langle A(a),x_o-x\rangle+f(x_o)\leq f(x)$$
We do not really need that $D$ is open, but for the reader who does not want to relearn separation Theorems in $\mathbb{R}^n$, knowing the proof of it in infinite dimesnional subspaces here is what one needs
Addendum: $D$ is convex and $f$ is continuous, hence $D$ is open and convex.
Indeed, if $(x_1,c_2),(x_2,c_2)\in D$ and $\alpha$ lies in $[0,1]$: $$f(\alpha x_1+(1-\alpha)x_2)\leq \alpha f(x_1)+(1-\alpha) f(x_2)<\alpha c_1+(1-\alpha)c_2$$ Therefore, we have that from the definition: $$(\alpha x_1+(1-\alpha)x_2, \alpha c_1+(1-\alpha)c_2)\in D$$
Now, let us prove that $f$ is continuous. This is a bit harder. First we need to prove that it is locally bounded.
To this end, given $x$, we have that if $z\in B_{\ell^\infty}^{1/4n}(x)$, then $z=x+\sum_i \lambda_i e_i$ where $|\lambda_i|<\frac{1}{4n}$. In particular, one may define $\beta$ such that $\beta+\sum \lambda_i=1$. It is clear that $\beta>3/4$ and so, that $\frac{\beta}{2n}+\lambda_i>\frac{1}{8n}>0$ and $\sum_{i=1}^n \left(\frac{\beta}{2n}+\lambda_i\right)+\sum_{i=1}^n \frac{\beta}{2n}=1$. More importantly: $$z=\sum_i \left(\frac{\beta}{2n}+\lambda_i\right)\left(x+ \frac{1}{2n}e_i\right)+\sum_i \frac{\beta}{2n}\left(x-\frac{1}{2n}e_i\right)$$ Now, by convexity, $$f(z)\leq \sum_i \left(\frac{\beta}{2n}+\lambda_i\right)f\left(x+ \frac{1}{2n}e_i\right)+\sum_i \frac{\beta}{2n}f\left(x-\frac{1}{2n}e_i\right)\leq$$ $$ \max_{i}\{f(x\pm e_i/2n)\}=:M$$ It is not difficult to see that $f$ is also bounded from below. Indeed, given $z$ in the neighborhood considered above:
$$f(x)=f\left(\frac{z}{2}+\frac{x+(x-z)}{2}\right)\leq \frac{f(z)+f(2x-z)}{2}$$
In particular, one has that $f(z)\geq 2f(x)-M$ as $x+(x-z)$ is easily seen to be in the prescribed neighborhood.
Now that we have that $f$ is locally bounded, we prove that $f$ is also locally Lipschitz. Suppose by contradiction that it isn't in $x$. Then one obtains a sequence of pairs $x_n,y_n$ such that $x_n,y_n\in B_{l^{\infty}}^{1/8n}(x)$ such that:
$$f(x_n)-f(y_n)> n^2 \|x_n-y_n\|$$
By connecting $y_n$ to $x_n$, we may send $x_n$ to a circular ring where the bound still holds and where we have complete mastery over its norm. Explicitly, we take $x_n=\alpha_n y_n+(1-\alpha_n)z_n$ where $\|z-x\|_{\infty}=\frac{1}{2}(\frac{1}{4n}+\frac{1}{2n})$ and $\alpha_n\in [0,1]$. By convexity:
$$\alpha_nf(y_n)+(1-\alpha_n)f(z_n)-f(y_n)\geq f(x_n)-f(y_n)> n^2 \|x_n-y_n\|_\infty$$
Rearranging and recalling that $x_n-y_n=(1-\alpha_n)(z_n-y_n)$
$$ f(z_n)-f(y_n)\geq n\|z_n-y_n\|_\infty\geq n^2 \left(\frac{1}{2}\left(\frac{1}{4n}+\frac{1}{2n}\right)-\frac{1}{8n}\right)=\frac{n}{8}$$
Because our sequence is bounded, we obtain $M-(2f(x)-M)\geq n/8$, which is a clear contradiction by letting $n$ increase.
