# Norm of an Operator

• Apr 19th 2010, 02:12 PM
ejgmath
Norm of an Operator
I need to show that the operator $\displaystyle T:L^2([0,1],\mathbb{R}) \rightarrow L^2([0,1],\mathbb{R})$ defined by $\displaystyle (Tf)(x)=xf(x)$ has norm 1

There is a hint on the question saying show it must be at least $\displaystyle 1-\frac{1}{n}$ for all $\displaystyle n$

The norm for $\displaystyle L^2$ is $\displaystyle \|f\|_2=\left(\int_{0}^{1}|f(x)|^2dx\right)^\frac{ 1}{2}$

Any help would be great
Thanks
• Apr 19th 2010, 06:17 PM
surjective
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.
• Apr 21st 2010, 09:14 AM
ejgmath
Okay, so I have shown $\displaystyle T$ is linear, pretty straighforward. I am stuggling with the bounded part.

I think I somehow need to show $\displaystyle \|Tf\|\leq R\|f\|$ for some $\displaystyle R>0$. Please correct me if I'm wrong.

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

• Apr 21st 2010, 12:44 PM
Opalg
Quote:

Originally Posted by ejgmath
I need to show that the operator $\displaystyle T:L^2([0,1],\mathbb{R}) \rightarrow L^2([0,1],\mathbb{R})$ defined by $\displaystyle (Tf)(x)=xf(x)$ has norm 1

There is a hint on the question saying show it must be at least $\displaystyle 1-\frac{1}{n}$ for all $\displaystyle n$

The norm for $\displaystyle L^2$ is $\displaystyle \|f\|_2=\left(\int_{0}^{1}|f(x)|^2dx\right)^\frac{ 1}{2}$

You know that $\displaystyle \|f\|_2^2 = \int_{0}^{1}|f(x)|^2dx$, and $\displaystyle \|Tf\|_2^2 = \int_{0}^{1}x^2|f(x)|^2dx$. Multiplying the integrand by $\displaystyle x^2$ has the effect of making it smaller, because $\displaystyle x^2<1$ throughout the interval except at the right-hand endpoint. That tells you that $\displaystyle \|Tf\|_2 < \|f\|_2$, 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 $\displaystyle f(x) = x^n$. 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 $\displaystyle \|f\|_2^2$ and $\displaystyle \|Tf\|_2^2$ for the function $\displaystyle f(x)=x^n$, you should find that $\displaystyle \frac{\|Tf\|}{\|f\|} = \sqrt{\frac{2n+3}{2n+5}}$, which may not be "at least $\displaystyle 1-\tfrac1n$" as the hint would like, but it serves the same purpose of being close to 1.
• Apr 21st 2010, 12:54 PM
surjective
bounded operator
Hello,

Would the following be correct:

$\displaystyle \Vert (Tf)(x) \Vert = \Vert xf(x) \Vert \leq \vert x \vert \cdot \Vert f \Vert$.

Since

$\displaystyle \max_{x\in [0,1]} x=1$

we see that:

$\displaystyle \Vert (Tf)(x) \Vert \leq 1 \cdot \Vert f \Vert$
• Apr 22nd 2010, 08:58 AM
Opalg
Quote:

Originally Posted by surjective
Hello,

Would the following be correct:

$\displaystyle \Vert (Tf)(x) \Vert = {\color{red}\Vert xf(x) \Vert \leq \vert x \vert \cdot \Vert f \Vert}$.

Since

$\displaystyle \max_{x\in [0,1]} x=1$

we see that:

$\displaystyle \Vert (Tf)(x) \Vert \leq 1 \cdot \Vert f \Vert$

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 $\displaystyle \|Tf\|\leqslant\|f\|$ is that if f, g are functions satisfying $\displaystyle 0\leqslant g(x)\leqslant f(x)$, for all x in the interval [a,b], then $\displaystyle \int_a^b\!\!\!g(x)\,dx\leqslant\int_a^b\!\!\!f(x)\ ,dx$. In this case, since $\displaystyle x^2|f(x)|^2\leqslant|f(x)|^2$ for $\displaystyle 0\leqslant x\leqslant1$, it follows that $\displaystyle \int_0^1\!\!\!x^2|f(x)|^2dx\leqslant\int_0^1\!\!\! |f(x)|^2dx$. Therefore $\displaystyle \|Tf\|^2\leqslant\|f\|^2$.