Edit: This question arose due to a misunderstanding, which has now been resolved.

Let $\psi \in L^{2}(\mathbb{R})$ be a continuous function. Let $M_{\psi}$ be the multiplication operator on $L^{2}(\mathbb{R})$. I need to find the domain of this operator. In his book, "Linear Operators on Hilbert Spaces'', Weidmann claims (on Page 52, Example 2) it to be the whole of $L^{2}(\mathbb{R})$.

We know that the multiplication operator has the full domain iff the function is essentially bounded. However, not all continuous square integrable functions are essentially bounded.

So is Weidmann making a mistake or am I?

For reference, let us construct a continuous, square integrable unbounded function.

First, define $f(x) = 0 ,\, x \leq 1$. Now let $n \in \mathbb{N}$. Define

$$ f(x) = n,\;\; \, x \in [n+1/4n^{3}, n + 1/2n^{3}] $$

Now linearly extend the function so that $f(n+1/2n^{3}) = 0$. Then let

$$ f(x) = 0 , \;\; x \in [n+1/n^{3}, n+1] $$

and extend linearly so that $f((n+1)+1/4(n+1)^{3}) = n+1$.


$$ \int_{\mathbb{R}}f \leq \sum\limits_{n=1}^{\infty}\frac{1}{n^{2}} < \infty $$

and hence $ f \in L^{2}(\mathbb{R})$. By construction it is continuous.


In example 2, page 52, Weidmann does not consider multiplication operators, but the functional $T \colon L^2(\mathbb R) \to \mathbb C$ $$ Tf = \int_{\mathbb R} \psi(x)f(x) \, dx $$ which of course is defined on the whole of $L^2(\mathbb R)$ for $\psi \in L^2(\mathbb R)$. It's just the scalar product with $\bar\psi$. The multiplication operator is defined on $L^2(\mathbb R)$ only for $\psi \in L^\infty(\mathbb R)$ as you say correctly.

  • $\begingroup$ Oh correct. A real blunder on my part. Actually the discussion above revolved too heavily with multiplication operator, so I oversaw the actual definition. $\endgroup$ – Vishal Gupta Aug 2 '14 at 11:15

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