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I have bigger problems with these 2 questions.

a) Let $\mathbf{X}$ be a Banach space and let $\mathbf{S:X→R}$ be linear and not bounded. Let $\mathbf{Y = graph(S) ⊂ X×R}$. Show that the map $\mathbf{T: Y → X}$ given by $\mathbf{T((x,Sx)) = x}$ is bijective and continuous, but not open.

b) Let $\mathbf{X}$ be a Banach space and let $\mathbf{S :X →R}$ be linear and not bounded. Let $\mathbf{Y =graph(S)}$ and define $\mathbf{F : X → Y}$ by $\mathbf{Fx = (x,Sx)}$. Show that $\mathbf{F}$ is closed, but not continuous.

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Equip $X \times \mathbb{R}$ with the norm $\left\|(x,r)\right\|_\infty = \max\{\|x\|,|r|\}$.

Clearly $T$ is linear. We have

$$\|T(x,Sx)\| = \|x\| \le \left\|(x,Sx)\right\|_\infty, \quad\forall x\in X$$

so $T$ is bounded and hence continuous.

To show that $T$ is bijective, notice that the inverse of $T$ is precisely $F$.

$F$ is not continuous. Namely, if $F$ were continuous, then $S = \pi_\mathbb{R}\circ F$ would be continuous, which it isn't. Here $\pi_\mathbb{R} : X \times \mathbb{R} \to \mathbb{R}$ is the projection $(x,r) \mapsto r$ which is known to be continuous.

If $T$ were open, $T^{-1} = F$ would be continuous, which is not true.

To show that $F$ is closed, assume $x_n \xrightarrow{n\to\infty} x$ in $X$ and $Fx_n = (x_n, Sx_n) \xrightarrow{n\to\infty} (z, Sz) \in Y$. Then clearly $x_n \xrightarrow{n\to\infty} z$ so $x = z$ and therefore $(z, Sz) = (x, Sx) = Fx$.

mechanodroid
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