I need to show that the operator defined by has norm 1

There is a hint on the question saying show it must be at least for all

The norm for is

Any help would be great

Thanks

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- April 19th 2010, 03:12 PMejgmathNorm of an Operator
I need to show that the operator defined by has norm 1

There is a hint on the question saying show it must be at least for all

The norm for is

Any help would be great

Thanks - April 19th 2010, 07:17 PMsurjective
Hello,

Try showing that the operator is linear. Then apply the definition of a bounded linear operator and the norm of the operator will naturally follow. - April 21st 2010, 10:14 AMejgmath
Okay, so I have shown is linear, pretty straighforward. I am stuggling with the bounded part.

I think I somehow need to show for some . Please correct me if I'm wrong.

I went along the lines of the cauchy schwarz inequality but I'm a bit stuck.

Thanks in advance - April 21st 2010, 01:44 PMOpalg
You know that , and . Multiplying the integrand by has the effect of making it smaller, because throughout the interval except at the right-hand endpoint. That tells you that , and the only hope of replacing that inequality by something approaching equality would by if the "mass" of the function f is concentrated close to 1. So try taking . For large values of n, that function will be very small through most of the interval and then shoot up to the value 1 when x=1.

If you caculate the integrals for and for the function , you should find that , which may not be "at least " as the hint would like, but it serves the same purpose of being close to 1. - April 21st 2010, 01:54 PMsurjectivebounded operator
Hello,

Would the following be correct:

.

Since

we see that:

- April 22nd 2010, 09:58 AMOpalg
No, the inequality in red is not justified. (The left side is a number, the right side is a function of x, which is 0 when x=0. But the left side is not 0.)

The reason that is that if f, g are functions satisfying , for all x in the interval [a,b], then . In this case, since for , it follows that . Therefore .