Show that there is $f_{0} \in \overline{\mathcal{F}}$ \such that $J(f_{0}) = \min_{f \in \overline{\mathcal{F}}}J(f)$

Question

Let $\mathcal{F} \subset C([0,1],\mathbb{R})$ be a family of functions such that:

• $f'(x)$ exist for every $x \in (0,1)$, $\forall f \in \mathcal{F}$;

• $\sup_{f \in \mathcal{F}}|f(0)| < \infty$;

• $M = \sup_{f \in \mathcal{F}}\sup_{x \in (0,1)} |f'(x)| < \infty$.

Let $J:= C([0,1],\mathbb{R}) \to \mathbb{R}$ the functional defined by $$J(f) = \int_{0}^{1}\sqrt{1 + f(x)^{2}}\mathrm{d}x.$$ Show that there is $f_{0} \in \overline{\mathcal{F}}$ such that $J(f_{0}) = \min_{f \in \overline{\mathcal{F}}}J(f)$, where $\overline{\mathcal{F}}$ denotes the closure of $\mathcal{F}$ relatively the norm of supremum.

My idea is to apply Ascoli-Arzelà to $\overline{\mathcal{F}}$ so, $\overline{\mathcal{F}}$ is compact and $J$ restrict to $\overline{\mathcal{F}}$ has a minimum. For this, I need to show that $\mathcal{F}$ is equicontinuous and equibounded. This is my problem.

Equicontinuity. Given $\epsilon > 0$, $|f'|$ is bounded, by the MVTheorem, $|f(x) - f(y)| \leq M|x-y|$. So, taking $\delta = \frac{\epsilon}{M}$ we conclude the result. Is correct?

Equibounded. I didn't get anything.

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Can someone help me?

Answer 1

For uniform boundedness, use $f(x)=f(0)+\int_0^x f'(y) dy$, then use your second and third hypotheses and the triangle inequality.