Two convex functions equal on the natural numbers are equal I have to prove that there is one and only one function $f \in C^1(\Bbb R_{>0})$ that satisfy the following. : 
$$\begin{aligned}f(x+1) - f(x) &= \ln(x)\\ \ \  \ f \rm{\ is\ convex}&\\f(1) &= 0\end{aligned}$$
I am not asked to find $f$ but just to prove its uniqueness. 
So far I thought of the following. Let $g$ that satisfy the set of equations above. Then $f = g$ on $\mathbb{N}$. So maybe the following is true : Two convex functions that are equal on the natural numbers are equal on $\Bbb R_{>0}$. Yet I am also unable to prove it or find a counterexample. 
 A: I think the claim is false, but I am not sure that my proof has no hole (if you find one, please let me know).
My approach is to create two different functions that will serve as the main ingredient to defining the derivatives of two convex functions which are a counterexample to your claim.
Take any two different, non-decreasing, integrable functions $f, g$ such that
$$ a = \int_0^1 \! f(x) \, \mathrm d x = \int_0^1 \! g(x) \, \mathrm dx \ne 0$$ 
and
$$ \frac{\mathrm d f(0)}{\mathrm dx} = \frac{\mathrm d f(1)}{\mathrm dx}, \frac{\mathrm d g(0)}{\mathrm dx} = \frac{\mathrm d g(1)}{\mathrm dx} $$
as well as
$$ \max_{x \in [0, 1]} f(x) = \max_{x \in [0, 1]} g(x), \quad \min_{x \in [0, 1]} f(x)  = \min_{x \in [0, 1]} g(x)$$
hold.
Now let $b$ denote the difference between the maximum value and the minimum value of $f$ (or $g$) on the interval $[0, 1]$ and define 
$$f^\ast(x) = b\lfloor x \rfloor + f(x - \lfloor(x) \rfloor) / a$$
and similarly $g^\ast$ to get our two monotonically non-decreasing functions serving as derivatives of the desired convex functions.
Then
$$ 
\begin{align*}
\int_n^{n + 1} \! f^\ast(x) \, \mathrm d x
 &= \int_n^{n + 1} \! b\lfloor x \rfloor \, \mathrm d x + \int_n^{n + 1} \! f(x - \lfloor(x) \rfloor) / a \, \mathrm d x \\
 &= bn + \int_0^{1} \! f(x) / a \, \mathrm d x \\
 &= bn + 1 \\
 &= \int_n^{n + 1} \! g^\ast(x) \, \mathrm d x.
\end{align*}
$$
Hence the functions
$$F(n) = \int_0^n \! f^\ast(x) \mathrm d x, \qquad G(n) = \int_0^n \! g^\ast(x) \mathrm d x $$
agree on all $n \in \mathbb N$ but not on all $x \in \mathbb R$.
So we reduced the problem to finding such $f, g$.
I think this is a bit tedious to write out, but one can use 
$$ f(x) = 1 + \sin(\pi x - \pi / 2)$$
which has
$$ \int_0^1 \! f(x) \, \mathrm d x = 1, \qquad \frac{\mathrm d f(0)}{\mathrm dx} = \frac{\mathrm d f(1)}{\mathrm dx} = 0$$
and
$$ 
g(x) =
\begin{cases}
0, & \text{if $x < 1/4$} \\
1 + \sin(2\pi x - \pi), & \text{if $1/4 \leq x < 3/4$ } \\
2, & \text{otherwise}.
\end{cases}
$$
which has the same discussed properties as $f$.
Using this, we obtain two convex functions $F, G$ which agree on $\mathbb N$ but not $\mathbb R$.
A: As Watercrystal shows, you cannot simply say that any two convex functions that agree on the integers are the same. However, with the condition $f(x+1)=f(x)+\log(x)$, makes the function unique. This is essentially the Bohr-Mollerup Theorem.

We will prove that this function exists and is unique for $1\le x\le2$, then the recursion $f(x+1)=f(x)+\log(x)$ proves the existence and uniqueness for all $x\gt0$.
Let $0\le x\le1$.
Convexity guarantees that
$$
\begin{align}
f(n+x)
&\le(1-x)f(n)+x\,f(n+1)\tag{1a}\\[6pt]
&=f(n+1)-(1-x)\log(n)\tag{1b}
\end{align}
$$
and
$$
\begin{align}
f(n+1)
&\le x\,f(n+x)+(1-x)f(n+1+x)\tag{2a}\\[6pt]
&=f(n+x)+(1-x)\log(n+x)\tag{2b}
\end{align}
$$
Therefore, with
$$
\begin{align}
\Delta_n(x)
&=\sum_{k=1}^{n-1}(\log(k+1)-\log(k+x))\tag{3a}\\
&=f(n+1)-f(n+x)+f(1+x)\tag{3b}
\end{align}
$$
$(1)$ and $(2)$ imply
$$
\Delta_n(x)-(1-x)\log(n+x)\le f(1+x)\le \Delta_n(x)-(1-x)\log(n)\tag4
$$
The lower limit in $(4)$ is increasing by Bernoulli's Inequality
$$
\begin{align}
&\Delta_{n+1}(x)-(1-x)\log(n+1+x)-\Delta_n(x)+(1-x)\log(n+x)\tag{5a}\\
&=\log\left(1+\frac{1-x}{n+x}\right)-(1-x)\log\left(1+\frac1{n+x}\right)\tag{5b}\\
&\ge0\tag{5c}
\end{align}
$$
and the upper limit in $(4)$ is decreasing by Bernoulli's Inequality
$$
\begin{align}
&\Delta_{n+1}(x)-(1-x)\log(n+1)-\Delta_n(x)+(1-x)\log(n)\tag{6a}\\
&=(1-x)\log\left(1-\frac1{n+1}\right)-\log\left(1-\frac{1-x}{n+1}\right)\tag{6b}\\
&\le0\tag{6c}
\end{align}
$$
The difference between the upper and lower limits in $(4)$ is decreasing to $0$
$$
\begin{align}
(1-x)(\log(n+x)-\log(n))
&\le\frac{x(1-x)}n\tag{7a}\\
&\le\frac1{4n}\tag{7b}
\end{align}
$$
Therefore, by the Squeeze Theorem and inequality $(4)$, there exists a unique $f(1+x)$ for $0\le x\le1$.
A: As I mentioned in my early comment, one can se that if $f$ is a function satisfying the conditions of the problem, than $F(x)=\exp(f(x))$ is Gamma like, that is 
$$F(x+1)=xF(x),\qquad  F(1)=1\tag{1}\label{one}$$
 It is a theorem by Bohr Mollerup that the only (continuous is enough) function that satisfies such a condition together with convexity of $f(x)=\log(F(x))$ is indeed the Gamma function. A proof by Artin though not terribly complicated is long. Starting with a function $F$ satisfying $\eqref{one}$ , one shows that
$$
\frac{F(x)}{F(1)}=\lim_{n\rightarrow\infty}\frac{n!n^{x-1}}{x(z+1)\cdot\ldots\cdot(x+n-1)}
$$
for all $0<x<1$. This gives existence and uniqueness. The steps to worked this out can be found as an exercise (problem 7, section 9.6) on Simon's Basic Complex analysis which is based on Artin's proof in his book, The Gamma function. I ignore whether the additional assumption ($f\in\mathcal{C}^1$) would streamline the arguments. 

Regarding the last statement of the problem, here is another counter example to the last statement in the problem.


*

*Consider any continuously differentiable and strictly convex function $f:[0,\infty)\rightarrow\mathbb{R}$, for instance $f(t)=t^2$.

*For each nonnegative integer $n$, define $g_n(t)=f'(n)(t-n)+f(n)$. This is the linear function tangent to $f$ at $t=n$. 

*Define $g(t)=\max_n g_n(t)$. This is a convex piecewise linear function, and $g(t)\leq f(t)$ with equality on at $t=n\in\mathbb{Z}_+$. 
The function $g$ is not continuously differentiable. There are several ways to smooth out the corners of $g$. Here is a geometric construction to do so.


*Two consecutive linear functions $g_n$ and $g_{n+1}$ intersect at a point $P_n(\xi_n,g(\xi_n))$ where $n<\xi_n< n+1$.

*Through the midpoints $M_n$ and $M_{n+1}$ of the segments $Q_n(n,f(n))$ to $P_n$ and $P_n$ to $Q_{n+1}(n+1),f(n+1))$ respectively,  draw orthogonal straight segments $\ell_n$ and $\ell_{n+1}$. These straight lines intersect at a point $C_n$, the orthocenter of the triangle $\triangle Q_nP_nQ_{n+1}$.

*Centered at $C_n$, draw a circular arc $\widehat{M_nM_{n+1}}$ the joining $M_n$ and $M_{n+1}$. This arc is tangent to the the graph of $g$ at the points $t_n=\frac{n+\xi_n}{2}$ and $t_{n+1}=\frac{\xi_n+n+1}{2}$. 

*It is not difficult to se how to define a function whose graph is obtained by glueing the segments $(Q_n,M_n)$ and the arcs $\widehat{M_nM_{n+1}}$.

