Let $(E,\Vert\cdot\Vert)$ be a normed vector space and $f:E \rightarrow \mathbb{R}$ be a twice Fréchet differentiable function with $\sup_{x \in E} \frac{\vert f(x)\vert}{1+\Vert x\Vert^3} < \infty$ as well as \begin{align} \Vert D^2f(x+h) - D^2f(x) \Vert_L < C_1 \Vert h\Vert \qquad \forall x,h \in E \end{align} for some constant $C_1$. For a multilinear form $B$ on $E$ we used here the notation $\Vert B \Vert_L:=\sup_{\Vert x \Vert=1} \vert B[x,\ldots,x]\vert $.
Why does this imply \begin{align} \Vert Df(x)\Vert_L < C_2 (1+\Vert x\Vert^2) \qquad \forall x \in E \end{align} for some constant $C_2$?
Let $y\in E$ arbitrary. Then $$ F(t)=Df(tx)(y), \quad t\in\mathbb R, $$ is a real valued differentiable function and $F'(t)=D^2f(tx)(x,y)$. Mean Value theorem implies that, there exists a $t^*\in (0,1)$, such that $$ Df(x)(y)-Df(0)(y)=F(1)-F(0)=F'(t^*)=D^2f(t^*x)(x,y) $$ Hence $$ \| Df(x)(y) \| =\|D^2f(t^*x)(x,y)+Df(0)(y)\|\le \|D^2f(t^*x)\|\|x\|\|y\|+\|Df(0)\|\|y\| $$ and thus $$ \| Df(x)\| \le \|D^2f(t^*x)\|\|x\|+\|Df(0)\| \\ \le \|D^2f(t^*x)-D^2f(0)\|\|x\|+\|D^2f(0)\|\|x\|+\|Df(0)\| \\ \le C_1 \|t^*x\|\|x\|+\|D^2f(0)\|\|x\|+\|Df(0)\| \\ \le C_1\|x\|^2+\|D^2f(0)\|\|x\|+\|Df(0)\| $$