Show that this linear operator is not continuous.

Question

This is a question of my last exam in Functional Analysis of my graduation:

Consider that norm: $|f|=\int_0^1 x^2 f(x)$ where $f\in C^0([0,1])$. Show that the linear operator $f(1-x)\in C^0([0,1])$ is not continuous.

MY ATTEMPT

I show that this operator is bounded with norm 1. So the operator is continuous and the teacher give me 0 in this question

MY FRIENDS ATTEMPT

Talking with my friends, they say that: "This norm is more "refined" that the L1 norm, so the operator is limited."

MY TEACHER BLA BLA BLA He say the operator is NOT limited but don't say the demonstration

I believe that all is wrong but I don't know how to show this...

@felipe: "$f(1-x) $" is not really notation that expresses a linear operator. What you want to say is that your operator maps $f\mapsto f\circ g$, where $g(x)=1-x$. Also, it is not "limited" but "bounded". — Martin Argerami, Apr 22 '14 at 14:15

score 3 · Answer 1 · answered Apr 22 '14 at 14:10

3

Write $T$ for your operator. If you let $f_n(x)=n(1-x)^n$, then $|f_n|\to0$, while $|Tf_n|\to1$.

answered Apr 22 '14 at 14:10

Martin Argerami

score 1 · Answer 2 · answered Mar 04 '17 at 09:43

There are 3 relevant spaces here. You need to be careful where each element is and what the relevant notion of convergence is.

The interval $[0,1]$
The Banach space $C[0,1]$ of continuous functions on the interval. This space has the norm $\|f\|=\int_0^1 x^2 |f(x)|dx$
The vector space $V$ of all automorphisms $C[0,1] \to C[0,1]$ with the operator norm. That is to say $\|A\|_{op} = \sup \{\|Af\| \colon \|f\|=1\}$.

Breaking down the definition of the operator norm we get. . .

$\displaystyle \|A\|_{op} = \sup \Big \{ \Big \|\int_0^1 x^2 |Af(x)|dx \Big\| \colon \int_0^1 x^2 |f(x)|dx = 1\}$.

What you want to prove is this value is undefined (infinite) for your given operator.

Your given operator is an element of $V$. That is to say a automorphism of $C[0,1]$ given by $Af(x) = f(1-x)$.

One strategy is to find a sequence of functions $f_n$ such that the RHS of the $:$ remains at $1$ while the LHS increases to infinity.

2 Answers2