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Let $X$ be a normed linear space of continuous real valued functions on $[0,1]$, with norm $$\lVert f\rVert = \max_{x \in [0,1]} \lvert f(x) \rvert$$ Let $g:[0,1] \to \mathbb{R}$ be defined by $$g(x)=\begin{cases} 0 , \text{if }x \in [0,1/2) \\ 1 , \text{if } x\in [1/2 ,1] \end{cases}$$ Show that the linear map $$T: X \to X,\quad f \mapsto gf$$ is continuous, and find its norm.

I am having trouble understanding how this linear map $T$ is being defined. We are taking a $f \in X$, that maps $[0,1] \to \mathbb{R}$ and we are multiplying it by this function $g$.

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The map $T$ is in fact not well-defined, since the product $gf$ need not be continuous. For instance, if $f$ is the constant function with value $1$, then $gf=g$ is not continuous. I'm guessing that the codomain of $T$ is intended to be not $X$ but a larger space, such as the space of all bounded functions $[0,1]\to\mathbb{R}$ with norm $\|f\|=\sup_{x\in[0,1]}|f(x)|$.

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  • $\begingroup$ Ah, you posted this immediately after I edited my question. Given this bigger space, how does this change the contradiction? $\endgroup$ – El Spiffy Jun 7 '17 at 1:28
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    $\begingroup$ @ElSpiffy. It disappears. $\endgroup$ – William Elliot Jun 7 '17 at 1:59
  • $\begingroup$ @ElSpiffy it still says "continuous" functions. What exactly did you edit? $\endgroup$ – Henno Brandsma Jun 7 '17 at 3:43
  • $\begingroup$ @HennoBrandsma appears it was edited off, I had added that if f was the constant function, then this would only hold if g was continuous. $\endgroup$ – El Spiffy Jun 7 '17 at 3:45
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    $\begingroup$ Perhaps you meant to restrict $X$ by requiring $f\in X\implies f(1/2)=0$? $\endgroup$ – DanielWainfleet Jun 7 '17 at 4:23

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