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, I know from this post that if $f'^2$ is integrable then $f$ is necessarily $\frac{1}{2}-$Hölder.
Thank you.
Note: all the integrals are taken in the Riemann sense.
Counterexample: Let $I_n$ be the interval $[n,n+1/2^n].$ On the first half of $I_n,$ define the function $g_n$ to be an isosceles triangle of height $n2^{n/2}.$ On the second half of $I_n,$ define $g_n$ to be the mirror image of the first triangle, but this time pointing downward. Set $g_n=0$ elsewhere. Then $g_n$ is continuous on $[0,\infty).$ We have
$$\int_{I_n}g_n = 0,\,\,\int_{I_n}|g_n| = \frac{n}{2^{{n/2}+1}}.$$
Now set $f_n(x) = \int_0^x g_n,$ $x\in [0,\infty).$ Then $f_n$ is supported in $I_n,$ is positive on the interior of $I_n,$ and peaks at the midpoint $n+1/2^{n+1}.$ Verify that
$$\frac{f_n(n+1/2^{n+1})-f(n)}{(n+1/2^{n+1}-n)^{1/2}}=\frac{n}{2^{n/2+2}}2^{n/2+1/2} = \frac{n}{2^{3/2}} \to \infty.$$
What this says is that the Holder $1/2$-norm of $f_n$ tends to $\infty$ as $n\to \infty.$ That's what we need.
So now let's put this all together. Define $g=\sum_{n=1}^{\infty} g_n$ and $f(x)= \int_0^x g.$ Then $f$ has all the properties needed for a counterexample. There's some details left to be checked, so please ask questions if you like.