Straight Line Segment is Shortest Path between Two Points

Theorem
Let $A$ and $B$ be points in a Euclidean space.

Let $\LL$ be the straight line segment lying on both $A$ and $B$.

Then $\LL$ is the shortest line that lies on both $A$ and $B$.

Proof
Let $A = \tuple {x_1, y_1}$ and $B = \tuple {x_2, y_2}$ be an arbitrary pair of points embedded in the Cartesian $\R^2$-plane.

Let $\LL$ be the straight line segment from $A$ to $B$.

We are to demonstrate that there does not exist a differentiable real function $f: \R \to \R$ such that:

and such that the arc length of the graph of $f$ from $A$ to $B$ is less than the length of $\LL$.

The proof relies on the Fundamental Theorem of Calculus and the definition of the the arc length of a differentiable real function.

there exists a real function $f: \R \to \R$ such that:
 * $\map f {x_1} = y_1$ and $\map f {x_2} = y_2$

and such that:
 * the arc length of the graph of $f$ from $A$ to $B$ is less than the length of $\LL$.

This means that $\map f x$ is continuous over $\closedint {x_1} {x_2}$ (and for the sake of the proof, differentiable over the same interval).

Thus, the arc length of the graph of $\map f x$ over $\closedint {x_1} {x_2}$ is:
 * $\ds L_f = \int_{x_1}^{x_2} \sqrt {1 + \paren {\map {f'} x}^2} \rd x$

Suppose that $L_f$ is less than the length of the line connecting $\tuple {x_1, y_1}$ and $\tuple {x_2, y_2}$.

We have that Slope of Curve at Point equals Derivative, which we will denote $s$.

Thus, the length of the line is given by:
 * $L_l = \ds \int_{x_1}^{x_2} \sqrt {1 + s^2} \rd x$

where $s = \dfrac {y_2 - y_1} {x_2 - x_1}$.

Since $L_f < L_l$, transitively, we have that:
 * $\ds \int_{x_1}^{x_2} \sqrt {1 + \paren {\map {f'} x}^2} \rd x < \int_{x_1}^{x_2} \sqrt {1 + s^2} \rd x$

Lemma $2$
Following from Lemma $1$ and Lemma $2$, we can reduce the inequality into:
 * $\ds \int_{x_1}^{x_2} \size {\map {f'} x} \rd x < \int_{x_1}^{x_2} \size s \rd x$

Since $s$ is constant, we can reduce the using the Fundamental Theorem of Calculus to:
 * $\ds \int_{x_1}^{x_2} \size {\map {f'} x} \rd x < \size {y_2 - y_1}$

There are now four possibilities:
 * $(1): \quad \map {f'} x$ is strictly positive over $\closedint {x_1} {x_2}$
 * $(2): \quad \map {f'} x$ is strictly negative over $\closedint {x_1} {x_2}$
 * $(3): \quad \map {f'} x$ is both negative and positive over $\closedint {x_1} {x_2}$
 * $(4): \quad \map {f'} x$ is $0$ over $\closedint {x_1} {x_2}$

The following argument applies to both Case $1$ and Case $2$ without loss of generality, and Case $3$ is dealt with similarly.

Case 1
For Case $1$, we have since $\size a = a$ for positive, real $a$:
 * $\ds \int_{x_1}^{x_2} \map {f'} x \rd x < y_2 - y_1$

Using the Fundamental Theorem of Calculus we have that:
 * $\map f {x_2} - \map f {x_1} < y_2 - y_1$

By our previous assumptions that $\map f x$ connects $\tuple {x_1, y_1}$ and $\tuple {x_2, y_2}$:
 * $y_2 - y_1 < y_2 - y_1$

which is a contradiction.

Case 2
For Case $2$, we have that $s$ must be negative.

Otherwise $\map f x$ is strictly decreasing over $\closedint {x_1} {x_2}$ but $y_2 > y_1$.

This would immediately contradict that $\map f {x_1} = y_1$ or $\map f {x_2} = y_2$.

Thus, we can flip both signs and apply the reasoning for Case $1$.

Case 3
With Case $3$, we have that since $\size a \ge a$ for all real $a$:
 * $\ds \int_a^b \size {\map f x} \rd x \ge \int_a^b \map f x \rd x$

From this the proof immediately follows from Cases $1$ and $2$.

Case 4
With Case $4$, the argument is applied using Case $1$, since $\size 0 = 0$.

All cases have been covered, and the result follows.