# How to Show a Functional Has Infinitely Many Minimizers?

I'm working on a problem in calculus of variations and need some help understanding why a particular functional has infinitely many minimizers. The problem is as follows:

Consider the set

$$\mathcal{M} = \{ y \in C^1([0, 1]) \mid y(0) = 0 \text{ and } y(1) = 1 \}$$

and the functional $$J: \mathcal{M} \subseteq C^1([0,1]) \to \mathbb{R}$$ given by

$$J(y) = \int_0^1 |y'(x)| \, dx.$$

I'm asked to show that $$J(y)$$ has infinitely many minimizers. Is it true that $$y(x) = x$$ is one such minimizer?

However, the problem states that there are infinitely many minimizers, and I'm not sure how to formally demonstrate this. Could anyone explain why this is the case and perhaps provide examples of other minimizers?

• You could start by considering only paths $y$ with $y' \geq 0$ and simplify $J$ for this case. Commented Aug 15 at 4:42

Let $$y \in \mathcal{M}$$. Then using the fundamental theorem of calculus: $$\int_0^1 y^\prime (s) ds = y(1) - y(0) = 1 -0=1.$$ This implies that $$1 = \int_0^1 y^\prime (s) ds \leq \int_0^1 |y^\prime (s) |ds = J(y)$$ by the monotonicity of the integral.

Therefore $$y$$ is a (global) minimizer of $$J$$ if $$J(y)=1$$.

You have already found that $$J(y) =1$$ if $$y(x) :=x$$, which makes the above an if and only if statement.

To show that there are infinitely many minimizers therefore boils down to finding a sequence $$(f_n)_{n \in \mathbb{N}}$$ of distinct functions in $$\mathcal{M}$$ for which $$J(f_n) =1$$ for all $$n \in \mathbb{N}$$.

Hint 1:

Consider $$J$$ applied to functions with non-negative derivative.

Solution:

If $$y \in \mathcal{M}$$ has non-negative derivative, then $$J(y) = \int_0^1|y^\prime (s) | ds= \int_0^1 y^\prime (s) ds =y(1) - y(0) = 1.$$ Therefore we are finished if we find a sequence of distinct functions in $$\mathcal{M}$$ with non-negative derivative (one such sequence is given in the second spoiler).

Solution Part 2:

Let $$n \in \mathbb{N}$$ and consider the function $$f_n : [0,1] \to \mathbb{R}$$ defined by $$f_n(x) := x^n$$. Then for all $$n \in \mathbb{N}$$: $$f_n \in \mathcal{M}$$, because $$0^n =0$$ and $$1^n =1$$. Of course they all have non-negative derivative and are distinct.

Noticing that due to the fundamental theorem of calculus, we have $$\int_0^1 y'(x)dx=1$$ for any admissible funcion $$y$$; and that $$\int_0^1|y'(x)|dx\geq 1.$$ (It is a basic result in multivariable calculus, treating y'(x) as a 1$$\times$$1 matrix and $$|\cdot|$$ a special case of matrix norm; it is intuitive anyway)

$$|y'(x)|$$ is integrable because $$y\in C^1$$ so $$|y'(x)|$$ is continuous.

Thus, any function with $$\int_0^1|y'(x)|dx= 1$$ will be a minimiser. This condition is satisfied by any weakly increasing function, including $$y(x)=x$$. (because their derivatives are always non-negative)

Another example will be $$y(x)=x^2$$.