Question

\(J:C_1([0,1])\rightarrow C_1([0,1])\) defined by \(J(f)(x)=\int_0^xf(t)dt\)
I need to find \(|||J|||\)

I have the definition \(|||L|||:=\text{sup}\{||L(x)|| \ : \ ||x||\le 1\}\). where \(L\) is a linear operation.

I am super confused on how to even start going about this....
does this not mean that \(|||J|||=\text{sup}\{||J(f)|| \ : \ ||f||\le 1\}\)
but then what is \(||f||\)?

does \(||f||=|||f|||\)?

Answer 1

@eliassaab

Answer 2

Do you mean [0,10] or [0,1]?
Do you mean D(f) or J(f)?

Answer 3

sorry edited

Answer 4

Suppose \(\max_{x\in[0,1]}|f(x)|\le 1$, then $f(x)\le1\) for \(x\in[0,1]\).
Thus \(\int_0^xf(t)dt\le\int_0^x1*dt=x\le 1\) for \(\forall \ x\in [0,1].\)
So, \(\max_{x\in[0,1]}|J(f)|\le 1\) and thus \(|||J|||\le1\)

Now let \(f(x)=1\), then \(f(x)\in C([0,1])\) and \(\max_{x\in[0,1]}f(x)=1\) and \(\max_{x\in[0,1]}|\int_0^x1*dt= 1\).
So \(\text{sup}\{\max_{x\in[0,1]}|\int_0^xf(t)dt| \ : \ \max_{x\in[0,1]}|f(x)|\le 1\}\ge 1\).
Thus we have that \(|||J|||=1\).

Answer 5

this is what I came up with, but not sure if I am way off or onto something.

Answer 6

its from a part in the book where it says its easy to show |||J|||=1. I hate when it says its easy, it never is.....\(\huge :(\)

Answer 7

sorry about the $ $ just imagine they are not there:)

Answer 8

You did it right.

Answer 9

oh wow, good deal

Answer 10

ty

Answer 11

YW

Answer 12

There is a small technical adjustment since f can be <0,
You should have said that
\[

|| J(f)||=\sup_{|x|\le 1}\left | \int_0^x f(t) dt \right|\le \\
\sup_{|x|\le 1}\int_0^x\left | f(t) \right|dt\le
\int_0^1\left | f(t) \right|dt\le \int_0^1 \sup_{t\in [0,1]}| f(t)| dt=\\
\sup_{t\in [0,1]}| f(t)|=||f||
\]

Answer 13

great thnx

Answer 14

YW

Answer 15

sorry im a little confused on where that would need to be, where did mine brake?

Answer 16

if \[
\int_0^xf(t)dt\le 1\]
this does not imply

\[
\left|\int_0^xf(t)dt\right|\le 1\]

Answer 17

It does if you follows my steps

Answer 18

oh yeah I see, good deal

Answer 19

So i have to show that the following linear functional is continuous on C([0,1]). L(f)=f(0).

So ||f||<=1 implies f(x)<=1 for all x in [0,1] implies f(0)<=1 all x in [0,1], so |||L|||<=1 so L is continuous, to show |||L||| =1 i just need to let f=1 then max f =1 showing |||L|||=1.

Is this one right?

Answer 20

\[
|L(f)|=|f(0)| \le ||f|\implies |||L|||<1\\
L(1)=1\implies |||L|||=1

\]

Answer 21

@eliassaab can you explain this step please

\(\large\int_0^1 \sup_{t\in [0,1]}| f(t)| dt=\sup_{t\in [0,1]}| f(t)|\)

TY in advance.

Answer 22

Let
\[
M=\sup_{t\in [0,1]}| f(t)| dt
\] then
\[

\int_0^1 M dt = M \int_0^1 dt = M(1)=M

\]

Answer 23

ahh sup|f(t)| is just a real number, i get it:)