Inequality proof. Bounded second derivative. Given $|f’’(x)| < c$ for all Real values of $x$ and a $c > 0$. Prove the following inequality:
$$| f(x+1) + f(x-1) - 2f(x)| < 2c $$
So I’m relatively sure it has something to do with the second difference formula. But I’m not entirely sure how to make sense of it.
Difference formula: $$ f’’(x) =  ( f(x + h) + f( x- h ) - 2f(x))/h.  (h -> 0)$$
 A: You can prove it with the 2nd difference formula but its easier to just work with MVT in my opinion. Lets say that $f$ is $C^2$ on an open set containing $[x-1,x+1]$. We have by MVT,
$$ f(x+1) - f(x) = f'(\xi_1), \\  f(x) - f(x-1) = f'(\xi_2), $$
for some $\xi_1 \in (x,x+1)$ and $\xi_2\in(x-1,x)$. So
$$ f(x+1) + f(x-1)-2f(x) = f'(\xi_1) - f'(\xi_2)$$
Applying MVT again, there is some $\xi_3\in(\xi_1,\xi_2)$ such that 
$$ f'(\xi_1)-f'(\xi_2) = f''(\xi_3)(\xi_1 - \xi_2)$$
Now as $\xi_1,\xi_2$ belong to an interval of length 2, their difference is at most 2. Thus
$$ |f(x+1) + f(x-1)-2f(x)| \le  |f''(\xi_3)||\xi_1 - \xi_2| \le 2c,$$
as required.
A: Thanks to user Peter for thanking me (nicomezi) for turning a rough draft into a readable answer.
By Mean Value Theorem, for some $t \in (x,x+1)$ :
$$f(x+1)-f(x)=f'(t).$$
Similarly, we have $s \in (x-1,x)$ such that :
$$-(f(x)-f(x-1))= -f'(s).$$
Replacing in the LHS of the inequality you want to prove and applying MVT again to $f'$, we have $u \in (x-1,x+1)$ so that :
$$|f'(t)-f'(s)|=|(t-s) f''(u)|.$$
But $t-s <2$, finally :
$$|f(x+1)+f(x-1)-2f(x)| < 2|f''(u)| <2c.$$
A: For fixed $x$ consider the function 
$$ 
g(t) = f(x+t) + f(x-t) \, .
$$
Taylor's formula gives
$$
 g(1) = g(0) + g'(0) + \frac 12 g''(t)
$$
for some $t \in (0, 1)$. Using  $g'(0) = 0 $ and  $|g''(t)| = |f''(x+t) + f''(x-t)|< 2c$ we get
$$
 |f(x+1) + f(x-1) -2f(x)| = |g(1) - g(0)| = \frac 12 |g''(t)| < c 
$$
which is a better result by the factor of $2$.
