Suppose x1 and x2 are two points that violate the lipschitz constant. In other words, the slope from x1,y1 to x2,y2 exceeds k. By the mean value theorem, f′(c) exceeds k for some c between x1 and x2. This is a contradiction, hence k is a valid lipschitz constant.
Continuing the above example, let u be an upper bound for the derivative of f, where f is continuously differentiable. Further suppose f has some lipschitz constant k < u. Let g = u-k, the gap between these values. Find a point s such that f′(x) is at least k+¾g. Select a neighborhood about s such that the derivative does not change by more than ¼g. Now the derivative throughout this neighborhood is at least k+½g. Take any two points x1 and x2 in this neighborhood. The lipschitz constant tells us the slope from x1,y1 to x2,y2 is no larger than k. Yet the mean value theorem implies a derivative smaller than k, when in fact all derivatives in this neighborhood are larger than k. This contradiction means the lipschitz constant can be no smaller than the largest derivative. When f is continuously differentiable, the lipschitz constant is precisely the least upper bound of the absolute value of f′(x).
This result generalizes to n dimensional space. This time we are interested in the largest directional derivative. When I move this way, f changes the fastest. This is determined by the maximum eigen value of the jacobian. Once again, if f is continuously differentiable on Rn, such that the derivative can be bounded, it has a lipschitz constant k that is precisely equal to the least upper bound of the maximum eigen value of the jacobian. The proof is just like the above; move in the direction of the corresponding eigen vector and apply the mean value theorem. I'll leave the details to you.