$\inf\int_{0}^{1} {f'(x)}^2 dx$, when $f: [0,1] →\Bbb R,\; f(0) = 0,\; f(1) = 1,\; \exists f' \;\text{and}\; f\;$' is continuous on [0,1]

Question

I am trying to

i) determine the infimum

ii) show that there's a function for which $\int_{0}^{1} {f'(x)}^2 dx$ is the infimum

iii) show if such function is unique.

I tried out several functions that suit the given condition, but couldn't see how $\int_{0}^{1} {f'(x)}^2 dx$ changes as $f(x)$ changes. How could we solve this problem?

Answer 1

Claim. Let $\,\mathscr X=\{\,g\in C^1[0,1]: g(0)=0\,\,\& \,\,g(1)=1.\}$. Then then functional $\,\varPhi(\,g)=\int_0^1 \big(\,g'(x)\big)^2\,dx$, attains a global minimum at $\,f(x)=x$, i.e. $$ \min_{g\in\mathscr X}\varPhi(g)=\varPhi(\,f)=1. $$

Proof. If $g\in\mathscr X$, then $g$ can be expressed as $g(x)=x+h(x)$, where $g\in C^1[0,1]$ and $g(0)=g(1)=0.$ We have that $$ \varPhi(g)=\int_0^1 \big(g'(x)\big)^2\,dx=\int_0^1 \big(1+h'(x)\big)^2\,dx=1+2\int_0^1\,h'(x)\,dx+\int_0^1 \big(h'(x)\big)^2\,dx \\=1+\int_0^1 \big(h'(x)\big)^2\,dx\ge 1=\varPhi(\,f), $$ since $\int_0^1 h'(x)\,dx=h(1)-h(0)=0$. Hence $\,\varPhi(g)\ge\varPhi(\,f)$, for all $g\in\mathscr X$.

Uniqueness. In particular, if $g(x)\not\equiv x$, then $h'\ne 0$, where $h(x)=g(x)-x$, and hence $\int_0^1\big(h'(x)\big)^2\,dx>0$, which implies that $\varPhi(g)>1=\varPhi(\,f)$. Thus $\varPhi$ possesses a unique minimiser.

Note. It can be shown with Calculus of Variation as well, but its quite complicated, if one wants a rigorous proof.

Answer 2

So we want to maximize $$J[y]=\int_{0}^{1} (y'(x))^2 \mathrm{d}x$$ We have that the Lagrange-function is $L(x, y, y')=(y')^2$, and we must have that $$\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{\partial L}{\partial y'}\right)=\frac{\partial L}{ \partial y}$$ And you just need to solve this differential equation.

Answer 3

You can solve this problem with variational calculus. Let $y=f(x)$ and $g(x,y,y')=(y')^2=(f'(x))^2$. Then the function $f(x)$ that gives infimum satisfies \begin{equation} \frac{\partial g}{\partial y}-\frac{d}{dx}\left(\frac{\partial g}{\partial y'}\right)=0, \end{equation} if we use the Euler–Lagrange equation. As $\cfrac{\partial g}{\partial y}=0$, \begin{align} &\frac{d}{dx}\left(2f'(x)\right)=0\\ &f(x)=ax+b \end{align} where $a,b$ are constant. Using the condition that $f(0)=0$ and $f(1)=1$, \begin{equation} f(x)=x\ . \end{equation}