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 !

  • 1
    $\begingroup$ Hint: take $f=1_{[a,1]}$ where $a < 1$ is a positive real number close to $1$. $\endgroup$ – Mindlack Jan 7 at 18:16
  • $\begingroup$ 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. $\endgroup$ – Dirk Jan 7 at 18:31

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 $1$.

  • $\begingroup$ 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 $T$. $\endgroup$ – GEdgar Jan 8 at 14:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.