TITLE: Solving inequality for convex functions.
QUESTION [6 upvotes]: A function $f : I \rightarrow \mathbb R$ on an interval $I$ is convex
  if $f((1-t)x+ty)\le (1-t)f(x)+tf(y), \forall x,y \in I, t \in [0,1]$
Assume now that $I$ is an open interval.  Show that if $f$ is convex
  then for each $c \in I$ there exists $m \in \mathbb R$ such that $m(x
 − c) + f(c) \le f(x)$ for all $x \in I$ , and if in addition $f$ is
  differentiable at $c$ then $f'(c)$ is the unique $m$ that works. In
  general, must m be unique?

(we have previously shown that convex functions are continuous)
I think I'm missing the point of the question.
Surely, by the mean value theorem
we can find some $y\in (x,c)$ such that $f'(y)=\frac{f(x)-f(c)}{x-c}$ setting $m=f'(y)$ we get $m=\frac{f(x)-f(c)}{x-c}, m(x-c)+f(c)=f(x)$ such an $m$ fits the inequality we're asked to show.
I also cannot see why $f'(c)$ would be the unique solution or why we would ever get a unique solution of the inequality at all for that matter.
Clearly I'm misunderstanding the question, so if someone could point out where I would appreciate that.
Thank you

REPLY [4 votes]: Here is an idea how you can prove it.

Take a point $a\in I$ to the left of $c$ and a point $b\in I$ to the right, i.e. $a<c<b$, and draw two secant lines: $\ell_a$ through $(a,f(a))$ and $(c,f(c))$ and $\ell_b$ through $(c,f(c))$ and $(b,f(b))$. Denote $k_a$ and $k_b$ their slopes respectively. 
By convexity, the graph of $f$ is under $\ell_a$ on $[a,c]$ and above it outside $[a,c]$. Similarly, the graph of $f$ is under $\ell_b$ on $[c,b]$ and above it outside $[c,b]$. It gives, in particular, that $k_a\le k_b$.
Let $a\to c^-$. The slope $k_a$ is monotonically increasing (again by convexity) and bounded from above by $k_b$, hence, converges to some value $M_1$. Similar argument when $b\to c^+$ gives that $k_b\to M_2$.
Since $k_a\le k_b$ we get $M_1\le M_2$. Now any $m\in[M_1,M_2]$ works.
If $f$ is differentiable at $c$ then $M_1=M_2$.

P.S. The interval $[M_1,M_2]$ is called subdifferential of $f$ at $c$.

In your argument, the point $y\in (x,c)$ you found by the mean value theorem depends on $x$, hence $m=f'(y)$ also does. However, the value of $m$ in the claim must be the same for all $x\in I$.