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I want to prove that for any $g \in C_0^\infty$, $g\colon[0,\infty) \times \mathbb R \rightarrow \mathbb R$, and $g = g(t,x)$,

$$\sup_{x\in \mathbb R} |g(t,x)| \leqslant \int_\mathbb R \hspace{2mm} \left(\left|g(t,x)\right| + \left| \frac{\partial g(t,x)}{\partial x} \right| \right) \hspace{2mm}dx$$

for any $t$.

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Oh, i'm sorry, g(x,t) should be change to g(t,x) above. – Misaj Apr 21 '12 at 8:15
up vote 1 down vote accepted

We have for all $t\in\mathbb R$ and $x\geq 0$, since $g$ has a compact support, that $$g(t,x)=\int_{-\infty}^x\frac{\partial g}{\partial x}(t,s)ds$$ hence $$|g(t,x)|\leq \int_{-\infty}^x\left|\frac{\partial g}{\partial x}(t,s)\right|ds\leq\int_{-\infty}^{+\infty}\left|\frac{\partial g}{\partial x}(t,s)\right|ds.$$ Since this bounded is independent of $x$ (and finite, since $(t,s)\mapsto \frac{\partial g}{\partial x}(t,s)$ still have a compact support, we are done.

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Really thank you . – Misaj Apr 21 '12 at 16:29

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