Is this sequence of function uniform convergence

Question

Let $C[0,1]=\{ f:[0,1] \to \mathbb{R} \mid f \text{ is continuous} \}$ equipped with sup norm $\lVert f \rVert_\infty = \sup_{x\in [0,1]} \lvert f(x) \rvert$. I don't know that a sequence function $f_n(x) = -nx+1$ in $(C[0,1], \lVert \cdot \rVert_\infty)$ is convergent or not. I try to show this sequence function is Cauchy sequence. Let $ \varepsilon >0 $. Choose $ N = $ _______ . Let $ m,n \geq N $. Then $\lVert f_n - f_m \rVert = \sup_{x\in[0,1]} \lvert f_n(x) - f_m(x) \rvert $ $$ \sup_{x\in[0,1]} \lvert f_n(x) - f_m(x) \rvert= \sup_{x\in[0,1]} \lvert (m-n)x \rvert = ?? $$ How should I do?

Answer 1

Since $f_n(1)$ diverges, this sequence can't even converge pointwise to any real function, not to mention uniform convergence.