Does there exist a shortest path which satisfies certain boundary conditions.

Question

For context, I saw a variant of this question posted a couple of days ago, however, the question seems to have been deleted. $$$$ Let $S$ be the set of $C^1$ functions $y$ where $y:\mathbb{R}\longrightarrow \mathbb{R}$ such that $y(0) = y(1) = 0$, $y'(0) = \alpha$ and $y'(1) = \beta$. $\alpha$ and $\beta$ are fixed and at least one of them is non-zero. Let $l:S\longrightarrow\mathbb{R}$, $l(f)=\int_0^1\sqrt{1+(f')^2} dx$. Does $L =\{l(f):f\in S \}$ have a minimum? $$$$ It is clear that $\inf(L) \geq 1$, however it does not seem like there should be a minimum of this set. I think this could be shown by first showing that $y = 0$ is the only function such that $l(y) = 1$ and $y(0) = y(1) = 0$-but clearly is not in $S$--then finding a subset $S'$ of $S$ such that $\inf(l(S')) = 1$, however, I was not able to accomplish this. Additionally would the question be different if each $y\in S$ was $C^{\infty}$?

Answer 1

A better result: Again let $\alpha =1,\beta =0.$ This time define $f_n(x) = x(1-x)^n.$ Then

$$f_n'(x)= (1-x)^n +xn(1-x)^{n-1}(-1).$$

The first summand is $\le 1$ in absolute value on $[0,1].$ So is the second summand, as a little work shows. Thus $|f_n'| \le 2$ for each $n.$ Furthermore, $f_n'(x)\to 0$ pointwise on $(0,1).$ By the DCT,

$$\int_0^1\sqrt {1+(f_n')^2} \to \int_0^1\sqrt {1+0} = 1. $$

The $f_n$ are polynomials, so we have improved on the previous result.

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Previous result: For convenience, let $\alpha =1,\beta =0.$ For $n=2,3,\dots $ define $f_n(x) = x(1-nx)^2,$ $x\in [0,1/n],$ $f_n=0$ on $[1/n,1].$ We have

$$f_n'(x) = (1-nx)^2 +x2(1-nx)(-n),\, x\in [0,1/n].$$

This shows $f_n'(1/n)=0,$ hence $f_n\in C^1$ for each $n.$ Note that $|f_n'|\le 3$ on $[0,1/n].$ Thus

$$\int_0^1\sqrt {1+(f_n')^2} \le \sqrt {3}\cdot (1/n) + 1-1/n.$$

The relevant infimum over $n$ is therefore $1,$ but the infimum cannot arise as $l(f)$ with the given constraints. Other values of $\alpha, \beta$ should be similar.