Is this operator continuous?

Question

Let $$ f(x) = \left\{ \begin{array}{ll} 0 & \mbox{if } x \leq 0 \\ -3 & \mbox{if } 0<x \leq 1 \\ 1/x & \mbox{if } x > 1 \\ \end{array} \right. \\ $$ and let the operator $T(g) = fg$ in the space $L_2(\mathbb{R})$.

I want to know if the operator is continuos and how I can get $\|T\|$.

What are or were your attempts? Pls enrich/edit your post correspondingly! Btw, $T$ looks rather continuous, and nice norm delimiters like in $|T|=3$ are obtained via the command "|". — Hanno, Jan 12 '20 at 12:41

score 2 · Accepted Answer · answered Jan 12 '20 at 12:39

2

Since $||f||_{\infty} = 3$, we get $||T(g)|| \leq 3||g||$. For $g = I_{[0,1]}$, the indicator function of $[0,1]$, we have equality. Hence $||T|| = 3$.

answered Jan 12 '20 at 12:39

Lukas Rollier

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This looks what i was looking for. And, what would be $ ker(T) $ ? – Miguel NoTeimporta Jan 12 '20 at 17:57
@MiguelNoTeimporta, any function that is $0$ a.e. on the positive real line. – Lukas Rollier Jan 13 '20 at 20:25

Is this operator continuous?

1 Answers1