I have the function $f: (0,1)^2 \to (0,+\infty)$ with type $$f(x,y) = x \log\left(\frac{x}{y}\right) + (1-x) \log\left(\frac{1-x}{1-y}\right)$$
I would like to get an upper bound for f(x,y) in terms of $|x-y|$.
I have the following idea. To simplify notation fix $y$ and let $g(x) = f(x,y)$. We have $$ g'(x) = - \log(\frac{1-x}{1-y}) + \log(\frac{x}{y})$$ and $$g''(x) = \frac{1}{1-x} + \frac{1}{x} = \frac{1}{(1-x)x}$$.
If $x < y$ using Taylor in the interval $[x,y]$ we have for a $\xi \in [x,y]$
$$ g(x) = g(y) + g'(y)(y-x) + g''(\xi) (y-x)^2/2 = g''(\xi) (x-y)^2/2. $$
If $x> y$ the using Taylor in the interval $[y,x]$ we get the same expression as above. Since $g''(x)$ is convex and is minimized at $x_0 = 1/2$ we have that if $|x-1/2| < |y-1/2|$ then $g''(\xi) \leq g''(y)$ else $g''(\xi) \leq g''(x)$.
Therefore, $$ g(x) = f(x,y) \leq \max(g''(x), g''(y))(x-y)^2/2 \leq (g''(x) + g''(y)) (x-y)^2/2 $$
Question Is the above proof correct? In particular I am not sure whether I can use Taylor in the closed interval $[x,y]$ or $[y,x]$.