Are $\phi$-contractions weak contraction? Let $(X,d)$ be a metric space and $f:X\longrightarrow X$. We recall:
(*) $f$ is said to be a weak contraction if $d(f(x),f(y))<d(x,y)$ for every $x,y\in X$ with $x\neq y$.
(**) $f$ is said to be a $\phi$-contraction if $d(f(x),f(y))\leq \phi(d(x,y))$ for certain  nondecreasing, upper semicontinuous function $\phi:[0,\infty)\longrightarrow [0,\infty)$ with $\phi(t)<t$ for every $t>0$.
On the other hand, in this paper (see Remark 2.3) is stated (without proof) that if $X$ is compact, then a weak contraction is a $\phi$-contraction. Do you think that is true of this statement?
Many thanks in advance for your comments.
 A: Define $\phi(t)=\sup \{d(f(x),f(y))|d(x,y)\le t\}$. Then, $\phi$ is obviously non-decreasing and $d(f(x),f(y))\leq \phi(d(x,y))$. Let's argue that $\phi$ is upper semi-continuous. 
Let $t\in [0,\infty)$ and $t_n\to t$ from above (which is the only case to check, since $\phi$ is non-decreasing). If $\limsup_{n\to\infty}\phi(t_n)>\phi(t),$ then there exists an $\varepsilon>0$ and sequences $x_n,y_n$ with $t<d(x_n,y_n)\leq t_n$ such that $d(f(x_n),f(y_n))\geq \phi(t)+\varepsilon$. Applying compactness, we can assume that $x_n$ and $y_n$ converge to $x$ and $y$ respectively. By continuity of $d$, $d(x,y)= t$. However, $d(f(x_n),f(y_n))\geq \phi(t)+\varepsilon\geq d(f(x),f(y))+\varepsilon$ which contradicts continuity of $f$. We conclude that $\phi$ is upper semi-continuous.
Now, clearly, $\phi(t)\le t$ since $f$ is a weak contraction. If $\phi(t)=t$ for any $t$, then there exists sequences $x_n,y_n$ such that $d(x_n,y_n)\le t$ and $d(f(x_n),f(y_n))\geq t-\frac{1}{n}$. Applying compactness again, we can assume that $x_n$ and $y_n$ are convergent with limits $x$ and $y$ respectively. However, by continuity, we get that $d(x,y)\le t$ and $d(f(x),f(y))=t$, which contradicts $f$ being a weak contraction. We conclude that $\phi(t)<t$ for every $t$ and you get the desired.
