Prove the following definitions for $\alpha$ are continuous

Question

Let $E = C[0,1]$ with the sup-norm. Prove that $\alpha: E \to \mathbb R$ in (i) and (ii) are Lipschitz. Prove that the function (iii) $\alpha:E \to E$ is continuous

(i) $\alpha(f) = f(\frac{1}{2})$

(ii) $\alpha(f) = \int_{0}^{1}f(x)dx$

(iii) $\alpha(f)(x) = 5xf(x)$

My attempts:

(i) By definition of the sup-norm, if $f,g \in E$ then $|f(1/2) - g(1/2)| \leq \text{sup}_{x \in [0,1]}|f(x) - g(x)| = ||f - g||$ so $\alpha$ is $1\text{-Lipschitz}$.

(ii) Again, if $f,g \in E$ then $|\int_0^1f - \int_0^1g| \leq \int_0^1|f-g| \leq \int_0^1||f-g|| = ||f-g||$ so $\alpha$ is again $1\text{-Lipschitz}$

(iii) Let $\epsilon >0$. If $||f-g|| < \frac{\epsilon}{5}$ then $$\text{sup}_{x \in [0,1]}|5xf(x) - 5xg(x)| = \text{sup}_{x \in [0,1]}|5x||f(x) - g(x)| \leq 5||f-g|| < \epsilon$$ which proves continuity. Are these correct, and if so, any improvements?

Answer 1

Yes, these are correct. No improvements are needed.

Answer 2

Yes these are correct, and pretty clear. In fact you have proven more than required for (iii); you proved $\alpha$ is uniformly continuous, which is even better.