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I am trying to find a function $f: \mathbb{R}^+ \to \mathbb{R}^+$ that fullfils the following conditions

  • $$f \in \mathcal{C}^1(\mathbb{R}^+,\mathbb{R}^+)$$

  • $$\int_{\mathbb{R}^+} f \in \mathbb{R}^+$$

  • $$\int_{\mathbb{R}^+} \mid f' \mid \in \mathbb{R}$$

  • $$f \text{is not $\frac{1}{2}$-Hölder}$$

I've tried functions with smooth spikes but I am unable to express this function as combinations of usual functions.

Moreover, It's worth noticing that if we have the assumption : $f'^2$ is integrable then $f$ is necessarily $\frac{1}{2}-$Hölder :

We have (using CS) : $$ \mid f(x) -f(y) \mid \leq \int_x^y 1 \times f' \leq \sqrt{\int_x^y f'^2}\sqrt{y-x} $$ Hence it follows that $f$ is $\frac{1}{2}-$Hölder continous since that $\sqrt{\int_x^y f'^2}$ is bounded.

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Indeed, disjoint tiny smooth spikes, of small heights and even much-much smaller widths, will do.

Let $K\in C^\infty(\mathbb R)$ be such that $K\ge0$, $K(x)=0$ if $|x|>1/2$, and $a:=K(1/3)-K(0)\ne0$. For instance, we may take $K(x)=\exp\frac1{4x^2-1}$ if $|x|<1/2$ and $K(x)=0$ if $|x|\ge1/2$.

Let $c:=\int_{\mathbb R} K<\infty$ and $b:=\int_{\mathbb R}|K'|<\infty$. For $j=2,3,\dots$, let $h_j:=1/j^2$, $d_j:=1/j^6$, and \begin{equation} f_j(x):=h_j K\Big(\frac{x-j}{d_j}\Big). \end{equation} Let \begin{equation} f:=\sum_2^\infty f_j. \end{equation} Then $f\in C^\infty(\mathbb R)$, $f\ge0$, \begin{equation} \int_{\mathbb R}f=\sum_2^\infty h_j d_j c<\infty, \end{equation} and \begin{equation} \int_{\mathbb R}|f'|=\sum_2^\infty h_j b<\infty. \end{equation} Yet, for $j=2,3,\dots$ \begin{equation} f(j+d_j/3)-f(j)=f_j(j+d_j/3)-f_j(j)=h_j a=a/j^2, \end{equation} which latter is much greater in absolute value than $\sqrt d_j=1/j^3$ as $j\to\infty$. So, $f$ is not $1/2$-Hölder-continuous.

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  • $\begingroup$ Great answer. It's quite impressive that adding the assumption : $f'^2$ is integrable prevent these kind of counter-examples. Do you have any ideas why (on an intuitive level) the integrability of $f'^2$ prevent the existence of a function with smooth spikes, of small heights and even much-much smaller widths which is not $1/2$-Hölder ? Thank you. (+1) $\endgroup$ – Thinking Nov 15 '18 at 12:00
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    $\begingroup$ @Thinking : Thank you for your comment. Concerning your question: This is a matter of scale. Indeed, in the example presented in my answer, given the height $h_j$, the derivative $f'_j\asymp h_j/d_j$ of the $j$th peak $f_j$ is inversely proportional to its "width" $d_j$ and hence large if $d_j$ is much smaller than $h_j$, but $|f'_j|$ is integrated over a small interval of length $d_j$, so that the integral $\int|f_j'|=h_j b$ is small (and does not depend on $d_j$). However, $(f'_j)^2$ is much greater than $|f_j|$, so that $\int(f_j')^2$ is on the order of $(h_j/d_j)^2d_j=j^2\to\infty$. $\endgroup$ – Iosif Pinelis Nov 15 '18 at 14:04
  • $\begingroup$ Indeed this make a lot of sense. Thank you for the detailed comment ! $\endgroup$ – Thinking Nov 15 '18 at 22:31

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