A function with only a partial derivative not Hölder-continuous

Question

I'm looking for a function of two variables, say $u(t,x)$, such that for some $\alpha\in ]0,1]$

1. $x\mapsto u(t,x)$ is $C^{2,\alpha}$;

2. $t\mapsto u(t,x)$ is $C^{1,\alpha}$;

3. $t\mapsto \partial_x u(t,x)$ is not $C^{0,\alpha}$.

All the statements must be true in a neighbourhood of $0$. I don't even now if a such a function can exists..

Thanks.

Answer 1

Let $\varphi:\mathbb R\to\mathbb R$ be a $C^\infty$-smooth compactly supported function such that $\varphi(0)=0$ and $\varphi'(0)\ne 0$. Define $$ u(x,t) = \begin{cases} \varphi(x/t),\quad & t>0 \\ 0,\quad & t\le 0\end{cases} $$

1. For every fixed $t$ the function $x\mapsto u(x,t)$ is $C^\infty$ smooth.
2. For every fixed $x$ the function $t\mapsto u(x,t)$ is $C^\infty$ smooth. (Indeed: if $x=0$, then it's zero identically. If $x\ne 0$, then $\varphi(x/t)=0$ when $t$ is sufficiently small, so transition at $t=0$ happens without a problem.)

Then take a look at the derivative $$ \partial_x u(x,t) = \begin{cases} t^{-1}\varphi'(x/t),\quad & t>0 \\ 0,\quad & t\le 0\end{cases} $$ and observe that

3. For $x=0$, the function $t\mapsto \partial_x u(x,t)$ has a serious problem at $t=0$.

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If the above example seems too extreme ($u$ is not even jointly continuous), you can use a milder one: $$ u(x,t) = \begin{cases} t\,\varphi(x/t),\quad & t>0 \\ 0,\quad & t\le 0\end{cases} $$

Answer 2

I have a curiosity about this examples! In some sense we assume that 1., 2. and 3. are not uniform in all variables. If we request a uniform condition, that is we say $x\to u(t,x)$ is $C^{0,\alpha}$ uniformly in $t$, if there exist a constant $C>0$ s.t. $$\sup_{\substack{{x,y\in\mathbb{R},{x\not=y}}\\t\in\mathbb{R}}}\frac{|u(x,t)-u(y,t)|}{|x-y|^\alpha}\le C.$$ In this case what can we say about the regularity of $t\to\partial_x u(t,x)$?

For instance in the second example, considering a positive sequence of point $x_n$ s.t. $x_n\to0^+$. If we compute the derivative in $t$ on the point $x_n$ we have $$\partial_t u(t,x_n)=\varphi(x_n/t)+\varphi'(x_n/t),$$ and this is not uniformly continuous in $x_n$:

• if we take $t=x_n$ we have convergence to $\varphi(1)+\varphi'(1)$ (in general different to zero),
• if we take $t=x_n^2$ we have convergence to zero.

Any hints, remarks and corrections are welcome!