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I am trying to prove that the distance between unilateral shift in $l^2$ and the set of normal operators is 1.

The desired distance is greater than or equal to 1 because of Zero operator.

In order to prove the distance is less than or equal to one, I showed that in the unit ball containing unilateral shift, there is no invertible operator.

Now, there is a hint that I should verify: "the set of invertible normal operator is dense in the set of normal operators".

Actually, I do not know how to verify that fact.

Moreover, how does it make sense to the main question.

Could you please help me a more detail hint?

Thank you so much.

Trần Linh
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    Left shift or right shift? The zero operator shows this distance is less than or equal to $1$, while the rest of what you say seems like an argument that it is greater than or equal to $1$. – David C. Ullrich Dec 13 '17 at 15:45

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My favorite version of the Spectral Theorem says that a normal operator is unitarily equivalent to a multiplication operator on some $L^2(\mu)$. That would be the operator $f\to mf$ for some $m\in L^\infty$.

It's clear that if $m\in L^\infty$ and $\epsilon>0$ then there exists $m'\in L^\infty$ with $|m'|\ge\epsilon$ almost everywhere and $||m-m'||_\infty\le\epsilon$.

David C. Ullrich
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