# Is this sequence of function uniform convergence

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?

• try to plug into $f_n(x)$ different values of $x\in [0,1]$ and compute the limits... – Avitus Sep 7 '15 at 13:29
• I have rarely seen a worse title. What does it mean, for a single function, to uniform converge? – Jack D'Aurizio Sep 7 '15 at 15:17
• Sorry. Please forgive me for my english skill. Now I have already edited my title. – C Pat Sep 7 '15 at 18:02

Since $f_n(1)$ diverges, this sequence can't even converge pointwise to any real function, not to mention uniform convergence.
• If I change the function to $f_n(x) = -nx+1$ when $x \in [0, \dfrac{1}{n}]$ and $f_n(x) = 0$ when $x \in [\dfrac{1}{n},1]$, the new function is also convergent or not. – C Pat Sep 7 '15 at 14:29