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.

  • 1
    $\begingroup$ T is not well defined. $\endgroup$
    – Ben W
    May 26, 2019 at 15:44
  • $\begingroup$ Please, how so @BenW? $\endgroup$
    – nandevers
    May 26, 2019 at 15:51
  • 2
    $\begingroup$ Linear maps are 0 at 0, not so for $Tx$ $\endgroup$ May 26, 2019 at 16:01
  • $\begingroup$ Your definition of $X$ yields $X = \mathbb{R}$... $\endgroup$
    – gerw
    May 26, 2019 at 17:52
  • $\begingroup$ Please include a complete quote of the question in Kreyszig because the sup norm over $\mathbb R$ does not norm $X$ $\endgroup$ May 27, 2019 at 14:23

2 Answers 2


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

  • $\begingroup$ Thanks, it’s tricky though. Because, to me it looked like a shift in the dominion of x and since x is defined on the whole of reals, it seemed ok. But indeed, the zero argument is very strong and obviously I missed it. $\endgroup$
    – nandevers
    May 26, 2019 at 16:10
  • $\begingroup$ I reviewed the quetion, and found that on Kreyszig it says that this is a linear and bounded operator. So I am reopening the question for newer insights on how to solve it. Page 102, Introduction to functional analysis, Problem 11, section 2.7. $\endgroup$
    – nandevers
    May 27, 2019 at 14:10
  • $\begingroup$ Well, there must be a mistake somewhere. $T$ is an operator which takes a linear operator and returns another linear operator. So $x(t)$ must be element of $X$, it is. However, $x(t-\Delta)$ must be also an element of $X$, which apparently isn't. $\endgroup$
    – Hume2
    May 27, 2019 at 14:25
  • $\begingroup$ Sure thing @Hume2, I’ll go over the details of both questions and try and see what the mistaken is. I`ll take the question down momentarily to avoid confusion. $\endgroup$
    – nandevers
    May 27, 2019 at 14:27
  • $\begingroup$ Apparently, I cant do that. So, I’ll just review it. $\endgroup$
    – nandevers
    May 27, 2019 at 14:30

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 $‖𝑥‖=\sup_{𝑡∈ℝ}|𝑥(𝑡)|$.

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. enter image description here

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 $0$.

  • $\begingroup$ Thanks a bunch. $\endgroup$
    – nandevers
    May 27, 2019 at 14:31
  • $\begingroup$ @upStoneLock yw! $\endgroup$ May 27, 2019 at 14:32

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .