Showing $h(t):=( s \mapsto\frac{s}{s+2+t})$ is continuous and calculating $\int_{-1}^1h(t)dt$.

Question

Let $C_b$ the set set of all continuous and complex functions on $[0,1]$ equipped with the uniform-norm (supremum-norm). Let $h:[-1,1] \rightarrow C_b[0,1]$ defined as $$h(t):=( s \mapsto\frac{s}{s+2+t})$$

I need to show that $h$ is continuous by showing that $||h(t_1)-h(t_2)||_\infty \rightarrow 0$ and calculate $\int_{-1}^1h(t)dt.$ The hint we received was that $T_s:C_b[0,1] \rightarrow \mathbb{C}, T(f) = f(s)$ is a bounded linear function for every $s \in [0,1]$.

For showing that $h(t)$ is continuous I thought about defining $t_2:=t_1 + \epsilon$. Then $||\frac{s}{s+2+t_1}-\frac{s}{s+2+t_2}||_\infty=||\frac{s}{s+2+t_1}-\frac{s}{s+2+t_1+\epsilon}||_\infty$ which goes to $0$ as $\epsilon \rightarrow 0$ thus $h$ is continuous but that seems too easy.

And how can I calculate $\int_{-1}^1h(t)dt$?

Thanks in advance!

Answer 1

if we denote $\int_{-1}^{1}h(t)dt$ by $f$ then $f(s)=\int_{-1}^{1}h(t)(s)dt$. (This is because $T_s$ is a continuous linear functional). Hence $f(s)=\int_{-1}^{1} \frac s {s+2+t}dt=s [\ln (s+3) -\ln (s+1)]$.