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If $x\in{\rm I\! R}^n$, then diagonal matrix $\mathop{\rm diag}(x)$ is a linear operator $\mathop{\rm diag}(x): {\rm I\! R}^n \to {\rm I\! R}^n$.

I am curious if there is some analogy for infinite dimensional space, like if $f\in C[0,1]$ then can we define somehow $\mathop{\rm diag}(f): C[0,1] \to C[0,1]$?

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what is diag(x) in R^n ? – Glougloubarbaki Nov 5 '12 at 20:14
It would be just multiplication, $(\text{diag}(f) (x))(t) = f(t)x(t)$. – copper.hat Nov 5 '12 at 20:16
Presumably $\text{diag}(x) (p) = \sum_k x_k p_k e_k$ – copper.hat Nov 5 '12 at 20:19
@copper.hat: You could write that as an answer. – joriki Nov 5 '12 at 20:26
@joriki: Thanks for the suggestion. – copper.hat Nov 5 '12 at 20:29
up vote 2 down vote accepted

For $f \in C[0,1]$, we could define $\text{diag}(f):C[0,1] \to C[0,1] $ by $(\text{diag}(f) (x))(t) = f(t)x(t)$. Basically pointwise multiplication.

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