Straight Line Segment is Shortest Path between Two Points

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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:

\(\ds \map f {x_1}\)

\(=\)

\(\ds y_1\)

\(\ds \map f {x_2}\)

\(=\)

\(\ds y_2\)

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

Note that for $x_1 = x_2$, a similar argument can applied to a function $\map f y$.

The validity of the material on this page is questionable. In particular: Note that this argument does not work if the points are such that $x_1 = x_2$. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by resolving the issues. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Questionable}} from the code. If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.

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

Aiming for a contradiction, suppose 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).

This article, or a section of it, needs explaining. In particular: How do we know that there is no real-valued function passing through $A$ and $B$ which is not differentiable? All this proof does is show that a differentiable function has the required property. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Explain}} from the code.

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 Tangent to Curve at Point equals Value of 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 $1$

Let $p, q \in \R$ be real numbers such that $\sqrt {1 + p^2} = \sqrt {1 + q^2}$.

Then:

$1 + \size p = 1 + \size q$

$\Box$

Lemma $2$

Let:

$\forall x \in \closedint a b: \map f x = \map g x$

Then:

$\ds \int_a^b \map f x \rd x = \int_a^b \map g x \rd x$

$\Box$

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 right hand side 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.

$\Box$

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 $1$.

$\Box$

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 $2$.

$\Box$

Case 4

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

$\Box$

When $x_1 = x_2$

When $x_1 = x_2$, we will take a function $f(y)$ to measure the arc-length instead of $f(x)$. From this alteration, which is distance preserving, we can apply our previous reasoning. It is distance preserving since we are taking the sum of $\sqrt{\d x^2 + \d y^2}$ instead of $\sqrt{\d y^2 + \d x^2}$, which are equivalent.

All cases have been covered, and the result follows.

$\blacksquare$