# What is the operator norm of $Tf(x) = x^2f(x)$?

Let $$H = L^2([0,1],\mathbb{R})$$ and $$T : H \to H,\, Tf(x) = x^2f(x)$$.

$$T$$ is linear.

$$\|Tf\|_{L^2([0,1],\mathbb{R})} = \sqrt{\int_0^1x^4f^2(x)dx} \leq\sqrt{\int_0^1f^2(x)dx} = \|f\|_{L^2([0,1],\mathbb{R})}$$

$$T$$ is linear and bounded therefore it's continuous. Also $$\|T|| \leq 1$$.

I tried finding solution to $$\|Tf\|_{L^2([0,1],\mathbb{R})} = \|f\|_{L^2([0,1],\mathbb{R})}$$

I found $$f(x) = \sqrt{\frac{2x-1}{x^4-1}}$$

but it's not in $$L^2$$ so it doesn't work.

anyone knows an $$f$$ to reach $$1$$, I'm not even sure it's $$1$$.

any help will be greatly appreciated !

• Hint: take $f=1_{[a,1]}$ where $a < 1$ is a positive real number close to $1$. – Mindlack Jan 7 at 18:16
• The answers below are fine, but one may add, that the approach by the OP can't work since there is no $L^2$ function for which the norm is attained. – Dirk Jan 7 at 18:31

## 2 Answers

Consider $$f_n(x)=x^n$$ for $$n\in\Bbb N$$.

Then $$\|f_n\|_2=\sqrt{\frac1{2n+1}}$$ and $$\|Tf_n\|_2=\sqrt{\frac1{2n+5}}$$ Since $$\lim_{n\to\infty}\frac{\|Tf_n\|_2}{\|f_n\|_2}=1$$ we have that $$\|T\|\ge1$$.

Multiplication by $$x^2$$ on $$[0,1]$$. $$T$$ is a self-adjoint operator. The essential range of the multiplier $$x^2$$ is $$[0,1]$$, so the spectrum of $$T$$ is $$[0,1]$$. The spectral radius of $$T$$ is $$1$$, and (since $$T$$ is self-adjoint), the norm of $$T$$ is $$1$$.

• As Dirk notes, there is no $f$ where the norm is attained. Here, we can see: $1$ is in the spectrum of $T$, but is not an eigenvalue of $T$. – GEdgar Jan 8 at 14:24