Proving a function is Lipschitz continuous Show that the following function is Lipschitz continuous and find a Lipschitz constant
$$y\mapsto f(x,y)\\
f(x,y)=\frac{y}{x}\ln(\frac{y}{x})\text{  , } |x-1|\leq\frac{1}{2}\text{  , } |y-e|\leq\frac{e}{2} $$
I have no clue on where to begin to prove this. 
My first question is how I should interpret this function, is $x$ a sort of constant?
 A: If $f$ is differentiable and the domain of definition $U$ is convex (which is true for the cartesian product of two intervals), then, for $x,y\in U$, by the mean value theorem, applied to 
$$\phi(t): = f(tx+(1-t)y)$$
there exists $t_0 \in (0,1)$ such that 
$$ f(x)-f(y) = \phi(1)-\phi(0)= \phi^\prime(t_0) =df(\xi)(x-y)$$
with $\xi :=t_0x +(1-t_0)y)$. Consequently, if you let $M:=\sup\limits_{ z\in U} ||df(z)||$, you get
$$|f(x)-f(y)|\le M |x-y|$$
Alternatively, e.g. if $U$ is not convex and $f\in C^1$, you can integrate: if $\gamma$ is a smooth curve joining $x$ and $y$, 
$$f(x)-f(y) = f(\gamma(1))-f(\gamma(0))= \int_0^1 (f\circ\gamma)^\prime(t) dt \le M\int_0^1||\gamma^\prime(t)||dt$$
with the same $M$ as before. 
The last integral is the length of $\gamma$. It depends on the geometry of $U$ whether, for each $x, y\in U$ you can find a curve $\gamma$ joining $x$ and $y$, s.t. $|x-y|\le C \,\mbox{length}(\gamma)$, with a constant $C=C(U)$. If this is true, then you get that $f$ is Lipshitz with Lipshitz constant $MC$.
A: I'm interpreting this problem as follows: You are given a family of functions
$$f_x(y):={y\over x}\log{y\over x}\qquad\left({e\over2}<y<{3e\over2}\right)\ ,$$
parametrized by $x$, ranging in ${1\over2}<x<{3\over2}$.
We are told to find a universal Lipschitz constant for each and every of these $f_x$. It will be sufficient to find such a constant for the same problem with all $<$ signs replaced by $\leq$.
When there is no need for an explicit "optimal" $L$, the existence of such an $L$ may be inferred from general principles concerning $C^1$-functions. Anyway, we shall construct in the following  an explicit Lipschitz constant $L$ for this problem.
By the MVT for any admissible $y_1$ and $y_2$ one has $$|f_x(y_2)-f_x(y_1)|=|f'_x(\eta)|\>|y_2-y_1|\ ,$$
where $\eta$ is admissible as well. It follows that
$$\max\left\{\bigl|f'_x(y)\bigr|\>\biggm|\>{e\over2}\leq y\leq{3e\over2},\ {1\over2}<x<{3\over2}\right\}$$
may serve as our Lipschitz constant $L$. One computes
$$f'_x(y)={1\over x}\left(1+\log{y\over x}\right)\ .$$
Looking at the rectangle $R$ of admissible points $(x,y)$ we see that ${e\over3}\leq {y\over x}\leq 3e$ on $R$, and it becomes obvious that $\bigl|f'_x(y)\bigr|$ is maximal when $x={1\over2}$, $\>y={3e\over2}$. Therefore we obtain
$$L={1\over{1\over2}}\bigl(1+\log(3e)\bigr)=2(2+\log 3)\ .$$
The $L$ so constructed is optimal for this problem: Let an $\epsilon>0$ be given. Choosing $x:={1\over2}+\delta$, $y_1:={3e\over2}-2\delta$, $y_2:={3e\over2}-\delta$ we have
$$f_x(y_2)-f_x(y_1)=f_x'(\eta)(y_2-y_1)>(L-\epsilon)(y_2-y_1)\ ,$$
if $\delta $ is choosen sufficiently small.
