Does Continuous diffentiablity imply Lipschitz continuity in Banach space?

Question

Let $X$ be a Banach space and $f:[0,\tau]\times X\to X$ is a continuously difference from $[0,\tau]\times X$ into X, then $f$ is continuous in t (first variable) and Lipschitz continuous in u (second variable) uniformly in $t$ on $[0;\tau]$. First part is clear. How do I prove second part? I got this question while studying the book '' Semigroups of linear operators and applications to partial differential equation'' by ''A pazy''.

Answer 1

We can show that $f$ is locally Lipschitz continuous on $u$ (uniformly in $t$), which means the following. For every $u\in X$, there is a neighborhood $N$ of $u$ and a constant $C$ such that $$||f(t,v)-f(t,u)||_X\leq C ||v-u||_X$$ for every $v\in N$ and $t\in [0,\tau]$.

Let $t\in [0,\tau]$ and $u\in X$ be fixed. Since $f$ is differentiable you may use the mean value theorem and get $$||f(t,v)-f(t,u)||_X\leq ||Df_X(t,\xi)\cdot(v-u)||_X$$ where $Df_X$ is the "partial" derivative in the $X$ direction and $\xi$ lies on the line segment between $u$ and $v$ in $X$.

Now $f$ is continuously differentiable so there's a neighborhood $N$ of $u$ in $X$ where $$||Df_X(t,\xi)||\leq ||Df_X(t,u)||+1$$ in the operator norm. This gives $$||f(t,v)-f(t,u)||_X\leq \big(||Df_X(t,u)||+1\big) ||v-u||_X$$ when $v\in N$. By compactness of $[0,\tau]$ we can get $$||f(t,v)-f(t,u)||_X\leq C||v-u||_X$$ where $C$ only depends on $u$, which is what we wanted to prove.