Straight Line Segment is Shortest Path between Two Points

Theorem
For any pair of points $\tuple {x_1, y_1}$ and $\tuple {x_2, y_2}$ in the $\R^2$-plane, there does not exist a real-valued, differentiable function $f(x)$ which connects $\tuple {x_1, y_1}$ and $\tuple {x_2, y_2}$ such that its length over $[x_1, x_2]$ is less than the straight line segment passing through both of them.

Proof
The proof relies on the Fundamental Theorem of Calculus and the formula for the arc length of any differentiable, real-valued function.

We start with a function $f(x)$ which "connects" $\tuple {x_1, y_1}$ to $\tuple {x_2, y_2}$.

This means that $f(x)$ is continuous over $[x_1, x_2]$ (and for the sake of the proof, differentiable over the same interval) and $f(x_1) = y_1$ and $f(x_2) = y_2$.

Thus, the length of $f(x)$ over $[x_1, x_2]$ is:
 * $L_f = \ds \int_{x_1}^{x_2} \sqrt{1 + \left[f'(x)\right]^2} \rd x$

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 $\ds s = \frac{y_2 - y_1}{x_2 - x_1}$.

Since $L_f < L_l$, transitively, we have that:
 * $\ds \int_{x_1}^{x_2} \sqrt{1 + \left[f'(x)\right]^2} \rd x < \int_{x_1}^{x_2} \sqrt{1 + s^2} \rd x$

Lemma 1
"p" We have that $p^2 = q^2$ implies that $|p| = |q|$ for real $p,q$.

This comes from the fact that $p^2 = |p|^2$ and since $p^2 \in \Z_{>0}$:
 * $\sqrt{p^2} = |p|$

Thus, $\sqrt{p^2} = \sqrt{q^2}$ by the injectivity of the square root.

Thus, $|p| = |q|$.

Since $1 + a$ is a bijection, $|p| + 1 = |q| + 1$.

Lemma 2
"$\ds \int_a^b f(x) \rd x = \int_a^b g(x) \rd x$ if $\forall x \in [a, b]: f(x) = g(x)$" Follows directly from the definition of the definite integral.

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

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

There are now three possibilities: The following argument applies to both Case 1 and Case 2 without loss of generality, and Case 3 is dealt with similarly.
 * $f'(x)$ is strictly positive over $[x_1, x_2]$
 * $f'(x)$ is strictly negative over $[x_1, x_2]$
 * $f'(x)$ is both negative and positive over $[x_1, x_2]$

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

Using the Fundamental Theorem of Calculus we have that:
 * $f(x_2) - f(x_1) < y_2 - y_1$

By our previous assumptions that $f(x)$ 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 $f(x)$ is strictly decreasing over $[x_1, x_2]$ but $y_2 > y_1$.

This would immediately contradict that $f(x_1) = y_1$ or $f(x_2) = y_2$.

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

Case 3
With Case 3, we have that since $|a| \ge a$ for all real $a$:
 * $\ds \int_{a}^{b} |f(x)| \rd x \ge \int_{a}^{b} f(x) \rd x$

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