Show that $\phi_x(t) = \int_{0}^{t}x(s)ds$ is continuous on $C[0,1]$

Question

Let $X=Y=C[0,1]$ with the uniform metric $d(x,y)=$sup$_{t\in [0,1]}\;\{|x(t) - y(t)|\}$

Define $$\phi:X \to Y$$

$$\phi_x(t) = \int_{0}^{t}x(s)ds$$ Show that $\phi$ is continuous on $X$

$d(\phi_x,\phi_y)=$sup$_{t\in [0,1]}\;\{|\int_{0}^{t}x(s)ds - \int_{0}^{t}y(s)ds|\}$=sup$_{t\in [0,1]}\;\{|\int_{0}^{t}(x(s)-y(s))ds|\}$

$|\int_{0}^{t}(x(s)-y(s))ds|\leq \int_{0}^{t}|x(s)-y(s)|ds$

How can I use this to prove it?

Answer 1

$\int_0^{t}|x(s)-y(s)|ds \leq d(x,y)$ so $d(\phi_x,\phi_y) \leq d(x,y)$.

Answer 2

I'd write $\phi(x)(t) = \int_0^t x(s)ds$. Then for any $x,y \in X$ and any $t \in [0,1]$:

$$|\phi(x)(t) - \phi(y)(t)| = |\int_0^t (x(s)-y(s))ds| \le \int_0^t |x(s)-y(s)| ds$$

and as for all $s$, $|x(s) -y(s)|\le d(x,y)$ we can estimate the final term above by $t d(x,y) \le d(x,y)$ (as $t \le 1$).

So know that for all $t$, $$|\phi(x)(t) - \phi(y)(t)| \le d(x,y)$$

and taking the sup on the left:

$$d(\phi(x), \phi(y)) \le d(x,y)$$ showing that $\phi$ is distance-shrinking and is even uniformly continuous: taking $\delta=\varepsilon$ will work at any point.