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Let $S$ denote the unit circle, so that $2\pi$-periodic functions can be treated as functions on $S$. For a given function $g\in C^1(S)$, define the multiplication map $M_g:C^1(S)\to C^1(S)$ by $(M_gf)(x)=g(x)f(x)$.

Prove that $M_g$ extends to a continuous linear map from $H^1(S)$ to $H^1(S)$, where $H^1(S)$ is the Sobolev space of order $1$, using that $\|f\|^2_{H^1}=\|f\|^2_{L^2}+\|\frac d{dx}f\|^2_{L^2}$.

My attempt: I have proven already that $C^1(S)\subset H^1(S)$, so I treat $M_g$ as a function from $C^1(S)$ into $H^1(S)$.

I can show that $M_g$ is linear, so my idea is to show that $M_g$ is also bounded, since that would imply continuity. So for $f\in C^1(S)$ I have $$ \|M_gf\|_{H^1}^2=\|M_gf\|_{L^2}^2+\left\|\frac{d}{dx}M_gf\right\|^2_{L^2}. $$ I need to find some $C$ so that $\|M_gf\|_{H^1}^2\leq C\|f\|_{H^1}^2$. I can bound the $\|M_gf\|_{L^2}^2$ by $$ \|M_gf\|^2_{L^2}=\int (g(x)f(x))^2=(\sup|g|)^2\int f(x)^2=(\sup|g|)^2\|f\|_{L^2}, $$ so that $\|M_gf\|_{L^2}^2\leq(\sup|g|)\|f\|_{L^2}$. But I'm not sure what to do with the $\|\frac d{dx}M_gf\|_{L^2}^2$.

Any help or hints on this problem would be greatly appreciated; thank you in advance.

Kimberly
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  • You can argue similarly, using Leibniz rule and extracting the sup of $|g|$ and $|dg/dx|$. – popoolmica Nov 13 '20 at 09:12
  • @popoolmica Hello and thank you for the tip - I have been trying to do that but am ending up with some extra terms. Do you mind providing a little more hint? – Kimberly Nov 13 '20 at 09:25
  • Sure. $\int\left|\partial_x \left(g(x)f(x)\right)\right|^2= \int \left| g'(x)f(x)+g(x)f'(x)\right|^2\le 2 \int \left| g'(x) f(x)\right|^2+2\int \left| g(x) f'(x)\right|^2$, and you can extract the two $\sup$'s. – popoolmica Nov 13 '20 at 10:00

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