# Norm of an Operator

• April 19th 2010, 03:12 PM
ejgmath
Norm of an Operator
I need to show that the operator $T:L^2([0,1],\mathbb{R}) \rightarrow L^2([0,1],\mathbb{R})$ defined by $(Tf)(x)=xf(x)$ has norm 1

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

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

Any help would be great
Thanks
• April 19th 2010, 07:17 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.
• April 21st 2010, 10:14 AM
ejgmath
Okay, so I have shown $T$ is linear, pretty straighforward. I am stuggling with the bounded part.

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

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

• April 21st 2010, 01:44 PM
Opalg
Quote:

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

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

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

You know that $\|f\|_2^2 = \int_{0}^{1}|f(x)|^2dx$, and $\|Tf\|_2^2 = \int_{0}^{1}x^2|f(x)|^2dx$. Multiplying the integrand by $x^2$ has the effect of making it smaller, because $x^2<1$ throughout the interval except at the right-hand endpoint. That tells you that $\|Tf\|_2 < \|f\|_2$, 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 $f(x) = x^n$. 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 $\|f\|_2^2$ and $\|Tf\|_2^2$ for the function $f(x)=x^n$, you should find that $\frac{\|Tf\|}{\|f\|} = \sqrt{\frac{2n+3}{2n+5}}$, which may not be "at least $1-\tfrac1n$" as the hint would like, but it serves the same purpose of being close to 1.
• April 21st 2010, 01:54 PM
surjective
bounded operator
Hello,

Would the following be correct:

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

Since

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

we see that:

$\Vert (Tf)(x) \Vert \leq 1 \cdot \Vert f \Vert$
• April 22nd 2010, 09:58 AM
Opalg
Quote:

Originally Posted by surjective
Hello,

Would the following be correct:

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

Since

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

we see that:

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

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