If $\,f$ and $g$ are continuous with compact support then $f*g$ (convolution ) is also continuous with compact support

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I do not know how to find a compact that satisfies $f$ and $g$ support. Could someone explain how find this new compact ?

How $f$ and $g$ are continuous in a compact then are bounded in this compact. But,

Are $f$ and $g$ bounded in all real line? Could someone starts the proofs?

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Compact support of $f*g$. Assume that $\,f,g\in C_0(\mathbb R)$. In particular, there exists an $M>0$, such that $f,g$ are supported in a subset of $[-M,M]$.

Then for $x>2M$, $$ (f*g)(x)=\int_{-\infty}^\infty f(t)\,g(x-t)\,dt=\int_{-M}^M f(t)\,g(x-t)\,dt=0, $$ since $x-t>M$, when $|t|\le M$ and $g$ vanishes for all such $x-t$. Similarly, $f*g$ vanishes for $x<-2M$.

Continuity of $f*g$.

Observe that $$ (f*g)(x+h)-(f*g)(x)=\int_{-\infty}^\infty f(t)\,\big(g(x+h-t)-g(x-t)\big)\,dt $$ and hence $$ |(f*g)(x+h)-(f*g)(x)|\le \max |f|\cdot \int_{-\infty}^\infty |g(x+h-t)-g(x-t)|\,dt $$ Then the right hand side tends to zero, as $h$ tends to zero, due to the uniform continuity of $g$. Note that since $g$ is continuous in $[-M,M]$, then it is uniformly continuous in the same interval.