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

    1. $(t,x)\mapsto \partial_x^2 u(t,x)$ is $C^{0,\alpha}$;
    2. $(t,x)\mapsto \partial_x u(t,x)$ is not $C^{0,\alpha}$.
    3. $t\mapsto u(t,x)$ is $C^{1,\alpha}$.

All the statements must be true in a neighbourhood of $(0,0)$. I'm beginning to think that such functions do not exist so the next question. If $u$ has a partial derivative $C^{0,\alpha}$, is $u\in C^{0,\alpha}$?

Thanks.

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1 Answer 1

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From conditions 1) and 3) it follows that $u(x,t)$ belongs to $C^{2+\alpha,1+\alpha}(\bar Q)$ for some cube $Q$. This space is a special case of anisotropic Besov spaces $B^{\mathbf s}_{\mathbf p\mathbf q}$. From the embedding theorem it follows (if I evaluated the exponents correctly) that $\partial_x u\in C^{1+\alpha,(1+\alpha)^2/(2+\alpha)}(\bar Q)$. Since $(1+\alpha)^2/(2+\alpha)>\alpha$ for $\alpha\in(0,1)$ functions satisfying 1)-3) do not exist indeed.

As for the last question, if $u$ depends only on $t$ then $\partial_x u\equiv0$ but $u$ can be nonsmooth.

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