Let $u\in L^\infty(\mathbb R\times (0,\infty))$ be a function such that $$u(x+z,t)-u(x,t)\leq c\left(1 + \frac 1 c\right)z\tag{$*$}$$ for some constant $c\in\mathbb R$ and almost all $x,z\in\mathbb R$, $t,z>0$. Let $\varepsilon>0$ and $\eta_\varepsilon\in C_C^\infty$ be the standard mollifier in the $x$ and $t$ variables with properties
- $\eta_\varepsilon\geq 0$
- $\operatorname{supp}\eta_\varepsilon\subset B_\varepsilon(0)$
- $\int_0^\infty\int_{\mathbb R} \eta_\varepsilon(x,t) dx dt = 1$
Then we have $$\frac{d}{dx}(\eta_\varepsilon*u)(x,t) \leq c'\left(1+\frac 1 t\right)$$ for all $t\geq 2\varepsilon$ for some constant $c'$ independent of $\varepsilon$.
I tried showing this, but I am not sure about my proof because it does not make use of $\varepsilon$, except in the mollifier $\eta_\varepsilon$:
From $(*)$ we have after dividing by $z$ and using $z\to 0$ that $$\frac{d}{dx}u(x,t)\leq c\left(1+\frac 1 t\right)$$ Now by partial integration we have $$\eta_\varepsilon*u_x(x,t) = \int_{\mathbb R} \eta_\varepsilon(x-y) u_x(y)dy = \frac{d}{dx}\left(\eta_\varepsilon*u\right)$$
and since $\eta_\varepsilon\geq 0$ we thus have the estimate \begin{align*} \frac{d}{dx}(\eta_\varepsilon*u) & = \int_{\mathbb R} \eta_\varepsilon(x-y)u_x(y)dy \\ & \leq c\left(1+\frac 1 t\right)\int_{\mathbb R} \eta_\varepsilon(z)dz \\ & \leq c\left(1+\frac 1 t\right)\int_0^\infty\int_{\mathbb R} \eta_\varepsilon dz dt \\ & = c\left(1+\frac 1 t \right) \end{align*}
Is this proof okay or otherwise can anyone help me correct it? Thanks!