Bounded Operator: Is D:L^2(0,1) Bounded?

In summary, there is no c>0 that satisfies the criteria for D to be a bounded operator, as shown by the example of defining f(x)=e^(dx) and observing that ||Df||=d^2||f||, which can be made arbitrarily large by choosing d>0.
  • #1
foxjwill
354
0

Homework Statement


Is the derivative operator [tex]D:L^2(0,1)\to L^2(0,1)[/tex] bounded? In other words, is there a c>0 such that for all [tex]f\in L^2(0,1)[/tex],
[tex]\|Df\|\leq c\|f\|?[/tex]​

Homework Equations


For all [tex]f\in L^2(0,1)[/tex],
[tex]\|f\| = \int_0^1 |f|^2\,dx.[/tex]​

The Attempt at a Solution


I'm pretty sure the answer is no. Here's my work:

Suppose [tex]c^2>0[/tex] satisfies the above requirements. Define [tex]f(x)=e^{(c+1)x}[/tex]. Then
[tex]\|Df\| = \int_0^1 (c+1)^2e^{2(c+1)x}\,dx = (c+1)^2\|f\| > c^2\|f\|.[/tex]​
But this contradicts the fact that [tex]\|Df\|\leq c^2\|f\|.[/tex] Thus, D is unbounded. Q.E.D.

Is this correct?
 
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  • #2
You have the right kind of example but the logic is a little overcomplicated. If you define f(x)=e^(dx) then ||Df||=d^2||f||. At this point you are done since d^2 can be chosen as large as you want. So there can't be a c such that ||Df||<c||f|| for any f.
 
  • #3
Dick said:
You have the right kind of example but the logic is a little overcomplicated. If you define f(x)=e^(dx) then ||Df||=d^2||f||.


No, just d||f|| ... provided d>0 ... but that is just as good.
 
  • #4
g_edgar said:
No, just d||f|| ... provided d>0 ... but that is just as good.

Right. There should have been a square root on the definition of ||f|| in the original post. Thanks for catching that.
 

1. What is a bounded operator in the context of D:L^2(0,1)?

In mathematics, a bounded operator is a linear transformation between two normed vector spaces that preserves the norm. In the context of D:L^2(0,1), it refers to a linear operator from the space of square-integrable functions on the interval (0,1) to itself.

2. How is the boundedness of an operator determined?

The boundedness of an operator is determined by its norm, which is defined as the supremum of the operator's values on a unit sphere. If this norm is finite, the operator is considered bounded.

3. What is the significance of bounded operators in functional analysis?

Bounded operators play a crucial role in functional analysis, as they provide a framework for studying linear transformations between normed spaces. They also have many important applications in areas such as differential equations, quantum mechanics, and signal processing.

4. How can the boundedness of an operator be proven?

To prove that an operator is bounded, one must show that its norm is finite. This can be done using various techniques, such as applying the Cauchy-Schwarz inequality or using the Banach-Steinhaus theorem.

5. What are the implications of D:L^2(0,1) being a bounded operator?

If D:L^2(0,1) is a bounded operator, it means that it is a well-behaved linear transformation that preserves the norm of square-integrable functions on the interval (0,1). This allows for the use of powerful analytical tools to study the operator and its properties.

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