# Is it possible to find $f$ such that : $f$ is absolutely integrable, $f'$ is absolutely integrable and such that $f$ is not $1/2$-Hölder

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.

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 $$$$f_j(x):=h_j K\Big(\frac{x-j}{d_j}\Big).$$$$ Let $$$$f:=\sum_2^\infty f_j.$$$$ Then $$f\in C^\infty(\mathbb R)$$, $$f\ge0$$, $$$$\int_{\mathbb R}f=\sum_2^\infty h_j d_j c<\infty,$$$$ and $$$$\int_{\mathbb R}|f'|=\sum_2^\infty h_j b<\infty.$$$$ Yet, for $$j=2,3,\dots$$ $$$$f(j+d_j/3)-f(j)=f_j(j+d_j/3)-f_j(j)=h_j a=a/j^2,$$$$ which latter is much greater in absolute value than $$\sqrt d_j=1/j^3$$ as $$j\to\infty$$. So, $$f$$ is not $$1/2$$-Hölder-continuous.
• 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) – Thinking Nov 15 '18 at 12:00
• @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 $(h_j/d_j)^2d_j=j^2\to\infty$. – Iosif Pinelis Nov 15 '18 at 14:04