Relation between uniform and operator norm Let $f:U\subset \mathbb{C}\to \mathbb{R}^n$ be a $\mathcal{C}^1$ function. I would like to know the relation between the uniform norm and the operator norm of the differential. For this question to make sense, I must make a few comments...
Remembering that the differential of the function $f$, is $f':U\to \mathcal{L}\left (\mathbb{C},\mathbb{R}^n\right )$. But in this special case, we are talking about a differentiable curve, and we have a natural isomorphism between $\mathcal{L}\left (\mathbb{C},\mathbb{R}^n\right )$ and $\mathbb{R}^n$ so the derivative can be seen as a function $f':U\to \mathbb{R}^n$. And so we have the uniform norm:$$\|f' \|_\infty =\sup \limits _{z\in U}\|f'(z) \|.$$Furthermore, seeing the differential as $f'\in \mathcal{L}\left (\mathbb{C},\mathbb{R}^n\right )$, we can apply the operator norm,$$\|f'(z)\|=\sup \limits _{|v|=1}\|f'(z)\cdot v\|.$$So, even with this information, I don't know if my question makes much sense, but I would like to know if the two norms are related, more specifically, if $\|f'\|_\infty <\infty$ we will also have $\|f'(z)\|<\infty$ for all $z\in U$.

What did I do:\begin{align*}\|f'(z)\| & =\sup \limits _{|v|=1}\|f'(z)v\|_{\mathbb{R}^n}=\sup \limits _{|v|=1}\|v(f'(z)\cdot 1)\|_{\mathbb{R}^n} \\
& "="\sup \limits _{|v|=1}\|v f'(z)\|_{\mathbb{R}^n} \\
& =\sup \limits _{|v|=1}|v|\|f'(z))\|_{\mathbb{R}^n}<\infty.
\end{align*}
Where in the second line I used the identification I mentioned earlier... I would like to know if my reasoning is correct...
 A: Let $f:U\subset \mathbb{C}\to \mathbb{R}^n$ being a $\mathcal{C}^1$ function.
Then $f(z)=(f_1(z),f_2(z),\ldots,f_n(z))$, in which $f_j:U\to \mathbb{R}$ are $\mathcal{C}^1$ functions. It follows that $f_j(x+iy)=u_j(x,y)$, in which $u_j:V\subset \mathbb{R}^2\to \mathbb{R}$ are $\mathcal{C}^1$ functions
This means that we can see $f$ as a $\mathcal{C}^1$ function $u:V\subset \mathbb{R}^2\to\mathbb{R}^n$. Therefore, we can see $f'(z)$ as a linear transform in $\mathcal{L}\left (\mathbb{R}^2,\mathbb{R}^n\right )$, or as a $n\times 2$ real matrix, or as a vector in $\mathbb{R}^{2n}$.
Let $$\|f' \|_\infty =\sup \limits _{z\in U}\|f'(z) \|_a$$ and $$\|f'(z)\|_{OP}=\sup \limits _{\|(v_1,v_2)\|=1}\|u'(x,y)(v_1,v_2)\|_b,$$ to some norm $\|\cdot\|_a$ in $\mathbb{R}^{2n}$ and some norm $\|\cdot\|_b$ in $\mathbb{R}^{n}$.
If $\|f' \|_\infty=M<+\infty$, then the absolute value of each coordinate of the Jacobian matrix $Ju(x,y)$, which represents the linear transform $u'(x,y)$, is bounded by some constant $0<N<+\infty$.
It follows that $$\|f'(z)\|_{OP}=\sup \limits _{\|(v_1,v_2)\|=1}\left\|Ju(x,y)\begin{bmatrix}v_1\\v_2\end{bmatrix}\right\|=\sqrt{s_1(f'(z))},$$ in which $s_1(f'(z))$ is the biggest eigenvalue of the matrix $Ju(x,y)^TJu(x,u)$. The Gershgorin circle theorem gives us a bound $0<P<+\infty$, in which $$0\leq s_1(f'(z))\leq P,\qquad z\in U.$$
Perhaps you can find related results searching for "\(\|A\|\) matrix norm" on SearchOnMath, for instance.
