Is Operator a Compact Operator?

Question

So I have this linear operator: $$ A : C_{[0,1]} \rightarrow C_{[0,1]}: (Ax)(t) = \int_{0}^{1}(ts + t^2s^2)x(s)ds $$ And I need to check if it is a compact one. According to Arzela-Ascoli theorem, I should check if $\operatorname{Im}(A)$ is precompact in $C_{[0,1]}$:
1)Is $\operatorname{Im}(A)$ uniformly bounded?
I should evaluate $|y(t)|$ such that $y \in\operatorname{Im}(A) $: $$ |y(t)| = \int_{0}^{1}(ts + t^2s^2)x(s)ds \le \left\| x \right\| \int_{0}^{1}(ts + t^2s^2)ds \le \left\| x \right\| \frac{3t+2t^2}{6} \le \left\| x \right\|$$ At this place I stopped. What should I do next? Or maybe I made a mistake somewhere?
2)Is $ \operatorname{Im}(A)$ equicontinuous?
By definition, $$ \forall y \in \operatorname{Im}(A): \forall \epsilon > 0 \exists\delta: \forall t_1,t_2 : |t_1 - t_2 | < \delta \Rightarrow |y(t_1) - y(t_2) | < \epsilon $$ Thus, I do next $$ |\int_{0}^{1}(t_1s + t_1^2s^2)x(s)ds - \int_{0}^{1}(t_2s + t_2^2s^2)x(s)ds| = |\int_{0}^{1}((t_1-t_2)s + (t_1^2-t_2^2)s^2)x(s)ds| < \delta\left\| x \right\| \int_0^1s+(t_1 + t_2)s^2ds = \delta \left\| x \right\| \frac{3 + 2(t_1+t_2)}{6} $$ And at this point I also stopped and have no clue what I should do next.
Hope for the help. Thanks.

Answer 1

Let $B$ be the closed unit ball (for the uniform norm) of $C[0,1]$. Then in order to establish uniform boundedness, it suffices to bound $\lVert x\rVert$ by one to see that all the elements of $\operatorname{Im}(B)$ have norm smaller than one.

For equicontinuity, you can continue the bound by writing $$ \left\| x \right\| \frac{3 + 2(t_1+t_2)}{6}\leqslant \frac{3 + 2(t_1+t_2)}{6}\leqslant 7/6, $$ since $t_1,t_2\leqslant 1$.

An alternative way to show compactness is that $(Ax)(t)=f_1(x)t+f_2(x)t^2$ where $f_1,f_2$ are linear continuous functionals on $C[0,1]$. Hence if $(x_n)$ is a sequence of elements of $B$, extract from $(f_1(x_n))$ a bounded subsequence $(f_1(x_{n_k}))$, and a further subsequence from $(f_2(x_{n_k}))$ to get a norm-convergent subsequence of $\left(Bx_n\right)$.