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Suppose $f(t,x)\in \mathcal{C}^0([0,1]\times \mathbb{R}^n)$. Further suppose that for each $t$ $$ C(t):= \sup_{x\in\mathbb{R}^n} |f(t,x)|<\infty \, .$$

Does it follow that $f$ is bounded?

Note that $C(t)$ need not be continuous as is evidenced by $f(t,x)= \sin(t x)$.

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    $\begingroup$ If $\mathcal C_0$ stands for continuous functions that vanish at infinity, they are all bounded even without further assumptions. If not, you need to clarify the notation. $\endgroup$
    – Emil Jeřábek
    Commented Sep 5, 2014 at 14:13
  • $\begingroup$ That was a typo. It has been fixed. $\endgroup$
    – Benji
    Commented Sep 5, 2014 at 14:15
  • $\begingroup$ To be explicit, what I mean is that $f$ is jointly continuous in $t$ and $x$. $\endgroup$
    – Benji
    Commented Sep 5, 2014 at 14:16

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It does not. Start from any non-zero continuous function $\phi$ with compact support in $\mathbb{R}\setminus\{0\}$ and define

$f(t,x)=\phi(tx)/t$ for $t>0$ and $f(0,x)=0$.

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    $\begingroup$ Note that the continuous dependence from $x$ is neither sufficient nor necessary for the positive answer to hold. If $\{f(\cdot, x)\}_x$ is an equicontinuous family of functions on $[0,1]$, then either $C(t)=\infty$ for all $t$ or $C$ is a continuous real valued function of $t$ (thus bounded). $\endgroup$
    – Pietro Majer
    Commented Sep 5, 2014 at 14:35

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