Show that this function is unique

Question

Let $(V,|.|_V)$ be a normed space and $(Y,|.|_Y)$ a Banach space. Suppose $X$ is dense in V and that for some $L>0$ the function $f:X\rightarrow{}Y$ satisfies

$$|f(x_1)-f(x_2)|_Y\le{L|x_1-x_2|_X}$$ for every $x_1,x_2\in{X}$.

I want to show there is a unique continuous function $F$ on $V$ that is equal to $f$ on X.

My attempt:

I have successfully done all of this apart from proving uniqueness. The function I used is:

$$\text{for }v\in{V},\text{define }F:V\rightarrow{Y},\text{ by }F(v) =\lim_{n\rightarrow{}\infty}{f(x_n)}$$ with $(x_n)$ a sequence converging to $v$.

Any hints on how to prove uniqueness would be appreciated.

Answer 1

Suppose that $F,G$ are two continuous functions such that $F=G=f$ on $X$. If $v\in V$ is arbitrary, pick $(x_n)\subset X$ such that $x_n\to v$. Then by continuity, $F(x_n)\to F(v)$ and $G(x_n)\to G(v)$, but $F(x_n)=G(x_n)=f(x_n)$, so, bu uniquness of limits on normed spaces, we have $F(v)=G(v)$, since $(f(x_n))$ converges both to $F(v)$ and $G(v)$.