Does $C^2_b$ regularity imply Lip. continuity with explicit constants?

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Assume $V,W$ are Banach. Let $C^2_b(V,W):=\{F:V\to W:\|F\|_{C^2_b}:=\|F\|_{\infty}+\|DF\|_{\infty}+\|D^2F\|_{\infty}<\infty\}$ where $D^nF$ is the $n$-th derivative of $F$. Is it true that $F$ and $DF$ are Lipschitz continuous with Lipschitz constants $\|F\|_{\infty}$ and $\|DF\|_{\infty}$ respectively?

I have been doing some reading and I believe they are since:

$|F(x)-F(y)|_W < |DF(b)|_{L(V,W)}|x-y|_V$ by the mean-value inequality since $V$ is convex and $|DF(b)|<|DF|_\infty$.

Is my reasoning accurate (carrying it over to the second derivative too)?

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Suppose $F \in C^1(V, W)$. Let $x, y \in V$ be arbitrary. Define $\phi : [0, 1] \to W$ by $\phi(t) = f(x + t(y - x))$. By the chain rule, $\phi'(t) = Df(x + t(y - x))(y - x)$, so $\phi \in C^1([0, 1], W)$. By the fundamental theorem of calculus (for the Bochner integral), $$f(y) - f(x) = \phi(1) - \phi(0) = \int_{0}^{1}\phi'(t)\,dt = \int_{0}^{1}Df(x + t(y - x))\,dt(y - x).$$ Thus $$|f(y) - f(x)| \leq \int_{0}^{1}|Df(x + t(y - x))|\,dt|y - x| \leq \sup_{t \in [0, 1]}|Df(x + t(y - x))| |y - x|.$$