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