Bounded linear operator $Tx(t) = x(t-\Delta)$?

Question

Let $X = (B(\mathbb{R},\mathbb{R}), \|.\|)$ be the space of all linear operators on $\mathbb{R} \rightarrow \mathbb{R}$ and define the operator $T:X \rightarrow X$ as $$Tx(t)= x(t - \Delta)$$ where $\Delta>0$ is fixed.

Is this operator linear? Is it bounded? And if so, what is its norm?

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To tackle this question, I came up with the following reasoning:

Take $x_1, x_2 \in X$, we know that they are them selves linear operators so all $x \in X$ is defined as $x:\mathbb{R} \rightarrow \mathbb{R}$ a linear and bounded operator.

Therefore, the application:

$T(x_1+x_2)(t) = (x_1+x_2)(t-\Delta)$

$T(x_1+x_2)(t) = x_1(t-\Delta) + x_2(t-\Delta)$

And $T(x_1+x_2)(t) = Tx_1(t) + Tx_2(t)$ so $T$ is linear.

As for boundedness, using the definition:

$\|Tx\| = \|x(t-\Delta)\|$

But since $x$ is itself bounded, for some real $c$:

$\|Tx\| = \|x(t-\Delta)\| \leq c\|(t-\Delta)\|$

And $T$ has the same norm as $x$, which I reckon should be the Euclidean norm.

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I hope you guys can shed some light on this and do criticize my arguments, please.

Some examples of such operator $T$ would be very nice too.

Answer 1

$x(t-\Delta)$ is not a linear operator because: $$x(0-\Delta) = -x(\Delta) \neq 0$$ So $T$ is not even an operator from $T: X \to X$.

Answer 2

this is a similar exercise to the one presented on Kreyszig’s Introduction to Functional Analysis, sec. 2.7, Problem 11. Exception that the norm is given as $‖‖=\sup_{∈ℝ}|()|$.

In the books solutions, the author says it is actually a linear operator and it is bounded.

I presume this is the question of Kreyszig, which is not in "Introduction to Functional Analysis", but rather "Introductory Functional Analysis with Applications". Note red box on why the two questions are different.

In particular note $\sup_{t\in\mathbb R}|x(t)|$ does not define a norm for $x\in B(\mathbb R,\mathbb R)$. The only element of $ B(\mathbb R,\mathbb R)$ with finite sup norm over $\mathbb R$ is $0$.