Let $f:X \to Y$ be a mapping between Hilbert spaces satisfying $$\frac{f(x+th)-f(x)}{t} \to f'(x)(h)$$ as $t \to 0$, i.e., $f$ is directionally differentiable.
If $f$ is known to be Lipschitz with Lipschitz constant $L$, is it true that $$L = \lVert{f'(x)}\rVert_{\text{operator}}$$ ?
The equality $L = \lVert{f'(x)}\rVert_{\text{operator}}$ does not look plausible, considering the right hand side involves $x$ while the left hand side does not. The correct statement is
$$L = \sup_{x\in X} \lVert{ f'(x) }\rVert_{\text{operator}}$$
Indeed, the inequality $\ge $ follows from $$\|f'(t)h\| = \lim_{t\to 0} \frac{\|f(x+th)-f(x)\|}{|t|} \le \frac{L \|th\|}{|t|} = L\|h\|$$ For the reverse inequality, let $M=\sup_{x\in X} \lVert{ f'(x) }\rVert_{\text{operator}}$. Let $\varphi$ be any unit-norm functional on $X$. It suffices to prove that the composition $g = \varphi\circ f$ is $M$-Lipschitz. Let $a,b\in X$ be distinct points, and consider the restriction of $g$ to the line segment $[a,b]$. This is a scalar function of a scalar argument, whose derivative is bounded by $M$. The mean value theorem yields $|g(a)-g(b)|\le M|a-b|$, which proves the claim.
Remark: This works for any pair of normed spaces, they don't have to be Hilbert spaces.