Show that a positive homogeneous uniform continuous map is Lipschitz

Question

Suppose $X$ and $Y$ are Banach spaces. Define a map $f:X \rightarrow Y$ which is uniform continuous and satisfies $$f(\lambda x) = \lambda f(x)$$ for $x \in X$ and $0<\lambda \in \mathbb{R}$.

Show that $f$ is Lipschitz continuous.

My attempt:

We want to show that $\| f(x) - f(y) \|_Y \leq M \| x-y \|_X$ for some constant $M>0$.

Note that $\| f(\lambda y) - f(\lambda x) \|_Y = \lambda \| f(x)-f(y) \|_Y$. How to continue from here? In particular, where should I apply uniform continuity?

If $X$ is compact, then I can use $\delta$ from uniform continuity repeatedly to obtain something. But I $X$ may not be compact.

Answer 1

Suppose that whenever $\|x-y\| \leq \delta$ then $\|f(x) - f(y)\| \leq \epsilon$.

Fix two points $X$ and $Y$ in $X$. Then, for some choice of $\lambda$, $\lambda X$ and $\lambda Y$ is such that $\|\lambda X - \lambda Y\| \leq \delta$. $\lambda = \frac{\delta}{\|X-Y\|}$ suffices.

But now, $\|f(X) - f(Y)\| = \frac{1}{\lambda} \|f(\lambda X) - f(\lambda Y)\| \leq \frac{\epsilon}{\lambda} = \frac{\epsilon}{\delta} \|X-Y\|$.