Csdrhrt

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 $f_n(x) = int_0^x g_n,$ $xin [0,infty).$ 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 $nto infty.$ 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.