The answer to the second question is also no. If $f$ is a Schwartz function, then $B_p\le C^p \| t^p\widehat{f}\|_{L^1}$, and this increases at most exponentially in $p$ if $\widehat{f}$ has compact support.

Let me also add some detail to what I said about the first part of your question in my comments above: We can build an $f=\sum f_n$, with $f_n$ supported near $x=n$, say, such that $|f^{(n)}(n)|\ge M_n\equiv (n+1)^{(n+1)n}$ (showing that $C_f=n$ doesn't work in your desired bound), but $f^{(j)}$ still stays bounded on $\mathbb R$ for each $j$.

To do this, take $f_n^{(n)}(x)= \pm M_n$, $\int f_n^{(n)}=0$, on a tiny symmetric interval $(n-\delta,n+\delta)$, or rather, take $f^{(n)}$ as a smooth version of this. The first property of $f_n$ is already built into this, and $f_n^{(j)}(x)$ for $j=0,1, \ldots , n-1$ can be kept as small as desired by taking $\delta>0$ sufficiently small.

Finally, we now see that we can also produce a compactly supported counterexample in just the same way.

The answer to the second question is also no. If $f$ is a Schwartz function, then $B_p\le C^p \| t^p\widehat{f}\|_{L^1}$, and this increases at most exponentially in $p$ if $\widehat{f}$ has compact support.

Let me also add some detail to what I said about the first part of your question in my comments above: We can build an $f=\sum f_n$, with $f_n$ supported near $x=n$, say, such that $|f^{(n)}(n)|\ge M_n\equiv (n+1)^{(n+1)n}$ (showing that $C_f=n$ doesn't work in your desired bound), but $f^{(j)}$ still stays bounded on $\mathbb R$ for each $j$.

To do this, take $f_n^{(n)}(x)= \pm M_n$, $\int f_n^{(n)}=0$, on a tiny symmetric interval $(n-\delta,n+\delta)$, or rather, take $f^{(n)}$ as a smooth version of this. The first property of $f_n$ is already built into this, and $f_n^{(j)}(x)$ for $j=0,1, \ldots , n-1$ can be kept as small as desired by taking $\delta>0$ sufficiently small, so if we keep those derivatives smaller than what we already had, then indeed $\|f^{(j)}\|_{\infty}\le \max_{0\le k\le j} \|f_k^{(j)}\|_{\infty}$ will be finite for each fixed $j$.

Finally, we now see that we can also produce a compactly supported counterexample in just the same way.

The answer to the second question is also no. If $f$ is a Schwartz function, then $B_p\le C^p \| t^p\widehat{f}\|_{L^1}$, and this increases at most exponentially in $p$ if $\widehat{f}$ has compact support.

The answer to the second question is also no. If $f$ is a Schwartz function, then $B_p\le C^p \| t^p\widehat{f}\|_{L^1}$, and this increases at most exponentially in $p$ if $\widehat{f}$ has compact support.

Let me also add some detail to what I said about the first part of your question in my comments above: We can build an $f=\sum f_n$, with $f_n$ supported near $x=n$, say, such that $|f^{(n)}(n)|\ge M_n\equiv (n+1)^{(n+1)n}$ (showing that $C_f=n$ doesn't work in your desired bound), but $f^{(j)}$ still stays bounded on $\mathbb R$ for each $j$.

To do this, take $f_n^{(n)}(x)= \pm M_n$, $\int f_n^{(n)}=0$, on a tiny symmetric interval $(n-\delta,n+\delta)$, or rather, take $f^{(n)}$ as a smooth version of this. The first property of $f_n$ is already built into this, and $f_n^{(j)}(x)$ for $j=0,1, \ldots , n-1$ can be kept as small as desired by taking $\delta>0$ sufficiently small.

Finally, we now see that we can also produce a compactly supported counterexample in just the same way.