Continuity of the multiplication map $f\mapsto x^2 f(x)$ between normed spaces

Question

Let $F:C[0,2]\to C[0,2]$ be the map defined by $(F(f))(x)=x^2f(x)$. Show that $F$ is continuous as a function from $(C[0,2],\|\cdot\|_{\sup})$ to $(C[0,2],\|\cdot\|_{2})$.

I read this solution:

Let $f\in C[0,2]$. Let $\epsilon>0$. Choose $\delta=\epsilon/(4\sqrt{2})$. If $\|g-f\|_{\sup}<\delta$ we have $$ \|F(g)-F(f)\|_2 = \left(\int_0^2 x^4(f(x)-g(x))^2\,dx\right)^{1/2} \le 2^2\sqrt{2}\|g-f\|_\sup<4\sqrt{2}\delta=\epsilon $$

I'm confused how the marker got the inequality from $$\left(\int_0^2 x^4(f(x)-g(x))^2 \ dx \right)^{1/2} \leq 2^2 \sqrt{2} ||g-f||_{sup}$$

Could someone please explain this step for me, and how they got the motivation to set $\delta = \epsilon/(4\sqrt{2})$?

Answer 1

As usual, $\delta$ was chosen after the estimate arrived at $\le 4\sqrt2 \delta$. For the estimate, note that $(f(x) - g(x))^2 \le \|g-f\|_\text{sup}^2$, thus $$\left(\int_0^2 x^4 (f(x)-g(x))^2 \ \mathrm dx \right)^{\frac12} \le \left( \|g-f\|_\text{sup}^2 \int_0^2 x^4 \ \mathrm dx\right)^{\frac12} = \sqrt{\frac{2^5}5} \|g-f\|_\text{sup}$$ So the constant could even be improved to $\sqrt{\frac{2^5}5} = 2^2 \sqrt{\frac25} < 2^2 \sqrt2$.

Answer 2

Note that if $f,g\geq0$ then we certainly have $$\int_a^bf(x)g(x)\ dx\leq(b-a)\max_{x\in[a,b]}f(x)\max_{x\in[a,b]}g(x).$$ If you're unsure of why, look at the graph. In this case, $x^4,(f(x)-g(x))^2\geq0$ and so $$\int_0^2x^4(f(x)-g(x))^2\ dx\leq2\cdot2^4\|f-g\|_\sup$$ from which your inequality follows by taking square roots. Usually in this kind of proof the choice of $\delta$ seems rather opaque until you work backwards. Choose $\delta>0$ and you will end up with $$2^2\sqrt2\|f-g\|_\sup\leq4\sqrt2\delta$$ and we want $4\sqrt2\delta\le\epsilon$, so $\delta=\frac\epsilon{4\sqrt{2}}$ is the natural choice.