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I would like to prove that on the space of absolutely continuous functions $AC([a,b])$ the following norms are equivalent: $$||f||_\# = ||f||_1 + ||f'||_1$$ $$||f||_{AC} = ||f||_{\infty} + ||f'||_1$$

One direction is trivial: $$||f||_1 = \int_{a}^{b}|f(t)|dt \leq (b-a)||f||_{\infty}$$ Hence: $$||f||_\# \leq max((b-a),1)||f||_{AC} = M||f||_{AC} \hspace{1em} \text{with } M \geq 0, \forall f \in AC([a,b])$$

I'm struggling on the proof of: $$\exists m\geq 0 \text{ s.t. } ||f||_{AC} \leq m ||f||_\# \hspace{1em} \forall f \in AC([a,b])$$ I've tried using foundamental theorem of calculus but I can't get anywhere. Do you have any idea?

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    $\begingroup$ There is a $c \in [a,b]$ with $\lvert f(c)\rvert \leqslant \frac{1}{b-a}\lVert f\rVert_1$. Use that to deduce $\lVert f\rVert_{\infty} \leqslant K\cdot \lVert f\rVert_{\#}$ for a suitable $K$. Then $\lVert f\rVert_{AC} \leqslant (K+1)\lVert f\rVert_{\#}$ follows. $\endgroup$ Commented Nov 13, 2019 at 21:52
  • $\begingroup$ I'm trying with mean value theorem combined with calculus formula but I can't really get to $||f||_\infty \leq K ||f||_\#$, can you give me an hint? $\endgroup$ Commented Nov 13, 2019 at 22:27
  • $\begingroup$ In particular when i write $||f||_\# \geq$ $...$ I don't find a way of getting $||f||_\infty$ on this side. $\endgroup$ Commented Nov 13, 2019 at 22:46

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On sets of positive measure the $L^1$-norm provides an upper bound for the infimum of the modulus of the function values. The proof is almost immediate.

Let $X$ be some space, $\mathscr{A}$ a $\sigma$-algebra on $X$, $\mu$ a positive measure on $\mathscr{A}$, $E\in \mathscr{A}$ with $0 < \mu(E) < +\infty$ (the case $\mu(E) = +\infty$ needs a small modification), and $\varphi \in L^1(\mu)$. Then $$\mu\biggl(\biggl\{ x \in E : \lvert \varphi(x)\rvert \leqslant \frac{\lVert \varphi\rVert_1}{\mu(E)}\biggr\}\biggr) > 0\,.$$

For otherwise we'd have $$\lVert \varphi\rVert_1 \geqslant \int_E \lvert \varphi(x)\rvert\,d\mu > \int_E \frac{\lVert \varphi\rVert_1}{\mu(E)}\,d\mu = \lVert \varphi\rVert_1$$ which is absurd.

In the case $\mu(E) = +\infty$ dividing by $\mu(E)$ would lead to $0$, and of course an integrable function generally need not vanish on a set of positive measure (if $\mu$ has atoms of infinite measure, then of course every integrable function must vanish (a.e.) on such an atom). But we can divide by any positive real number and obtain $\mu(\{x \in E : \lvert \varphi(x)\rvert \leqslant \varepsilon\}) > 0$ for every $\varepsilon > 0$. In fact, since $\mu(\{ x \in E : \lvert \varphi(x)\rvert > \varepsilon\}) < +\infty$ if $\varphi$ is integrable, we even have $\mu(\{x \in E : \lvert \varphi(x)\rvert \leqslant \varepsilon\}) = +\infty$ in this situation.

For the case at hand, this means that for every $f \in AC([a,b])$ there is a $c \in [a,b]$ with $$\lvert f(c)\rvert \leqslant \frac{\lVert f\rVert_1}{b-a}\,.$$ The set of such $c$ even has positive measure, but we only need one. Then, by absolute continuity of $f$, for every $x \in [a,b]$ we have \begin{align} \lvert f(x)\rvert &= \biggl\lvert f(c) + \int_c^x f'(t)\,dt\biggr\rvert \\ &\leqslant \lvert f(c)\rvert + \int_{\min \{c,x\}}^{\max\{c,x\}} \lvert f'(t)\rvert\,dt \\ &\leqslant \lvert f(c)\rvert + \int_a^b \lvert f'(t)\rvert\,dt \\ &\leqslant \frac{\lVert f\rVert_1}{b-a} + \lVert f'\rVert_1 \\ &\leqslant \max \biggl\{ \frac{1}{b-a}, 1\biggr\}\cdot \lVert f\rVert_{\#} \end{align} and thus $\lVert f\rVert_{\infty} \leqslant K\lVert f\rVert_{\#}$ with $K = \max \bigl\{ \frac{1}{b-a},1 \bigr\}$, which yields $$\lVert f\rVert_{AC} = \lVert f\rVert_{\infty} + \lVert f'\rVert_1 \leqslant K\lVert f\rVert_{\#} + \lVert f'\rVert_1 \leqslant (K+1)\lVert f\rVert_{\#}.$$

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  • $\begingroup$ Thank you for the detailed answer! For the first part isn't enough to use the integral mean value theorem? Since $f$ is continuous $\exists c \in [a,b] s.t. |f(c)| = \frac{||f||_1}{b-a} $. $\endgroup$ Commented Nov 14, 2019 at 19:13
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    $\begingroup$ Yes, for $f \in C([a,b])$ one can argue with the mean value theorem. But I wanted to emphasise that this holds in more general situations. $\endgroup$ Commented Nov 14, 2019 at 19:40

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