Let $f$ be a real-valued function with domain $[0,1]$. If there is some $K > 0$ such that $f(x) - f(y) \le K|x-y|$ for all $x,y \in [0,1]$, which of the following must be true?
I don't feel like typing up all the options, so I am just going to provide the answer. What follows from the above statement is that $f$ is continuous on $(0,1)$ , and according to a solution manual I found on UCLA, it is continuous because it is Lipschitz continuous. Aren't we missing some absolute value bars? I thought Lipschitz continuity was about $|f(x)-f(y)|$ being no larger than $K|x-y|$ for some $K > 0$, not $f(x)-f(y)$.
So, either the GRE question has an error or the solution has an error, which is it? Note: I tried to show that the above condition implies continuity, but I couldn't see how this could be done.
No error; as Kelenner noted in a comment, the assumption $$ f(x)-f(y) \le K|x-y| \quad \forall x\forall y \tag1$$ is equivalent to $$ |f(x)-f(y)| \le K|x-y| \quad \forall x\forall y \tag2$$ which is Lipschitz continuity. The point is that $$ |f(x)-f(y)| = \max(f(x)-f(y) , f(y)-f(x) ) $$ and $(1)$ bounds both terms on the right.