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Let $H=L^2([0,1],\lambda)$, where $\lambda$ is the Lebesgue measure on $[0,1]$. Let $f\in C([0,1])$ and consider the multiplication operator $M_f\colon H\to H$, $M_f(g)=fg$.

  1. Let $f,g\in C([0,1])$ such that $M_f=M_g$. I want to show that $f=g$.

It is: $M_f=M_g$ $\iff$ $fh=gh$ for all $h\in H$ $\iff$ (f-g)h=0 for all $h\in H$ $\iff$ $M_{f-g}(h)=0$ for all $h\in H$.

How to proceed without knowing that $M$ is isometric?

  1. Let $M_f$ be invertible as a linear bounded operator on $H$. I want to show that $f(x)\neq 0$ for all $x\in [0,1]$. I started as follows:

Assume that there exists an $x\in [0,1]$ such that $f(x)=0$. For $n\in\mathbb{N}$ consider the nonempty set $S_n=\{y\in [0,1]: |f(y)|<\frac{1}{n} \}$, and thus the nonzero characteristic funtion $\chi_{S_n}\in H$ regarding $S_n$. It is $$\|M_f(\chi_{S_n})\|_H=\|f\chi_{S_n}\|_h\le ||f_{|S_n}\|_{\infty}\|\chi_{S_n}\|_H\le \frac{1}{n}\|\chi_{S_n}\|_H.$$ Can I conclude that $M_f$ is not invertible from here and if yes, how?

Thank you

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    For the first point what happens if you take $M_{f-g}(f-g)$ ?

    For the second one if $M_f$ is invertible what information about $|M_f^{-1} |$ can you extract from your last inequality ?

     – Delta-u Apr 18 '18 at 17:22
  • ah right, thank you. The first point is now clear: $(f-g)^2=0$ and since $f-g$ is continuous, it follows that $f=g$. –  Apr 18 '18 at 17:25
  • @Delta-u regarding the second one, if I use this inequality https://math.stackexchange.com/questions/525970/norm-of-an-inverse-operator-t-1-t-1 , then it follows that $|M_f^{-1}| must be unbounded, right? This would contradict that it is invertible –  Apr 18 '18 at 17:31
  • Yes the idea is to show that for all $n$, $|M_f^{-1}| \geq n$ which is a contradiction. – Delta-u Apr 18 '18 at 17:39
  • thank you, @Delta-u! –  Apr 19 '18 at 06:50

2 Answers2

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The answer can be found here, and essentially we can consider the operator $T_{f}(g)=fg$ for $g\in C_{0}^{\infty}(0,1)$ and looking for $T_{f}=0$ to imply that $f=0$ a.e. on $(0,1)$. Since $f$ is continuous on $[0,1]$, we have $f=0$.

user284331
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  1. If $M_f = M_g$, then $f = M_f 1 = M_g 1 =g$ a.e. Because $f,g$ are assumed to be continuous, then $f=g$ everywhere because they are equal on dense subset of $[0,1]$.

  2. Suppose $M_f$ has a continuous inverse $L \in \mathcal{L}(L^2[0,1])$. Then, for $x\in [0,1]$ and $\epsilon > 0$, \begin{align} \|\chi_{[x-\epsilon,x+\epsilon]}\|&=\|LM_f\chi_{[x-\epsilon,x+\epsilon]}\| \\ & \le \|L\|\|M_f \chi_{[x-\epsilon,x+\epsilon]\cap[0,1]}\| \\ & \le \|L\|\cdot\sup_{[x-\epsilon,x+\epsilon]}|f|\cdot \|\chi_{[t-\epsilon,t+\epsilon]}\|.\end{align} Hence, it would follow that $1 \le \|L\|\sup_{[x-\epsilon,x+\epsilon]}|f|$ for all $\epsilon > 0$ and for all $x\in [0,1]$, from wich it follow that the continuous function $f$ cannot vanish anywhere in $[0,1]$.

Disintegrating By Parts
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