Let $V = C_\mathbb{R}[0,1]$, define $L \in L(V,V)$ as $(Lf)(x) = xf(x)$. Equip $V$ with the $L^2$-norm. I want to compute the operator norm $$ \|L\|_{B(V,V)} := \sup\{\|Lf\|_{L^2} : \|f\|_{L^2} \leq 1\} $$ And I have to show that for any $0 \neq f \in V$ we have have $\|Lf\|_{L^2} < \|L\|_{B(V,V)} \|f\|_{L^2}$.

What I have tried so far. To show $L \in B(V,V)$, I computed: $$ \|Lf\|_{L_2}^2 = \int_0^1 |xf(x)|^2dx = \int_0^1 |x|^2|f(x)|^2dx \leq \int_0^1 |f(x)|^2dx = \|f\|_{L_2}^2. $$ Thus, we have $\|Lf\|_{L_2} \leq k \|f\|_{L_2}$ for $k=1$, and $L$ is bounded. It also follows that $\|L\| \leq 1$.

I expected that we can show somehow that $\|L\| \geq 1$ by showing that there is some sequence $g_n \in C_\mathbb{R}[0,1]$ for which the norm converges to $1$, but I didn't manage to do this, I'm not sure if that's correct, someone has any ideas on how to do this?

  • $\begingroup$ Observe that if your function "focusses all its weight" near 1, then the inequality is almost an equality. Using this intuition perhaps we can find a sequence of functions of the type you require. $\endgroup$ – RandomStudent Mar 4 '19 at 18:42
  • $\begingroup$ Thank you for the hint. I don't know how to exploit this fact in a counter-example. I already tried to use the function $g: x \mapsto 1$, but this has a norm of $\frac{1}{2}$. $\endgroup$ – Sigurd Mar 4 '19 at 18:47
  • $\begingroup$ No, no, I wasn't referring to a counter example. Since you know that $||L|| \leq 1$, as you mentioned, all you need is a sequence of functions $f_n$ such that $||f_n|| = 1$ and $||Lf_n|| \to 1$. Such functions can probably be constructed using functions with weight concentrated near 1, that is something which is 0 "away" from 1. $\endgroup$ – RandomStudent Mar 4 '19 at 18:49
  • $\begingroup$ Yeah I see sorry I used the wrong terminology there. I'm interested in your idea but I'm still not sure how to construct such functions. $\endgroup$ – Sigurd Mar 4 '19 at 19:02


To reduce calculations, the following answer is probably way more tractable.

Pick $f_n(x) = \sqrt{2n + 1}x^n$. This "concentrates mass" near $1$. Also, trivially, $\norm{f_n} = 1$.

Also $\norm{Lf_n} = \displaystyle\int_{0}^{1} (2n + 1)x^{2n + 2} = \dfrac{2n + 1}{2n + 3} \to 1$ as $n \to \infty$ as required.

  • $\begingroup$ Feel free to accept if you're satisfied. :) $\endgroup$ – RandomStudent Mar 4 '19 at 19:33
  • $\begingroup$ Yes, this is a very nice one, thanks! :D How did you find that haha. $\endgroup$ – Sigurd Mar 4 '19 at 19:42
  • $\begingroup$ Lol, I was thinking about easy functions concentrated near $1$, and it suddenly occurred to me that the easiest functions I know satisfy that. :) $\endgroup$ – RandomStudent Mar 4 '19 at 19:43
  • 1
    $\begingroup$ It definitely is :) I'm already sick of the linear stuff and the integral calculations, this seems almost too smooth to be true :P $\endgroup$ – Sigurd Mar 4 '19 at 19:45

$\newcommand{\a}{\alpha}\newcommand{\b}{\beta}$ $\newcommand{\norm}[1]{\left\lVert#1\right\rVert}$ Consider functions of the form $f(x) = \a \dfrac{\sqrt{x - a}}{\sqrt{1 - \b}}$ when $x \geq \b$, otherwise $f(x) = 0$.

By calculation, $\norm{f} = \dfrac{\a^2}{1 - \b}\dfrac{(\b - 1)^2}{2}$. To set this to $1$, use $\a^2 = (1 - \b)\dfrac{2}{(\b - 1)^2}$.

Also $\norm{Lf} = \dfrac{\a^2}{1 - \b}\dfrac{\b^4 - 4\b + 3}{12}$. With the above restriction of choosing $\a, \b$ such that $\norm{f} = 1$, the expression for the norm is $\norm{Lf} = \dfrac{\b^4 - 4\b + 3}{12}\dfrac{2}{(\b - 1)^2} \to 1$ as $\b \to 1$.

Now, formally, we need to pick a series $f_n$ in your function space such that $\norm{f_n} = 1$ and $\norm{Lf_n} \to 1$. To do that set $f_n$ to be the $f$ with $\b = 1 - 1/n$, and $\a$ set as above so that $\norm{f_n} = 1$. Now, since $\b \to 1$, as $n \to \infty$, $\norm{Lf_n} \to 1$, as required, and so the operator

Note: This is definitely not the nicest function sequence, but yes, it (probably?) works.

  • $\begingroup$ Please note that the function sequence I refer to uses the idea I mentioned above, that the weights of the functions concentrate near $1$ as $n \to \infty$, while keeping the norm fixed at $1$. $\endgroup$ – RandomStudent Mar 4 '19 at 19:18
  • $\begingroup$ Thanks for your explanation! It is indeed rather complicated but that seems to work at first sight. I'm also trying to construct something based on your previous comment, namely to let $f_n$ be the linear interpolation from $1-\frac{1}{n}$ to $1$ on the x-axis and $0$ to $2n$ on the $y$-axis. This has norm 1 and it focusses the weight around $1$, but I still need to see if this works out with the norms and all. $\endgroup$ – Sigurd Mar 4 '19 at 19:21
  • $\begingroup$ Yes, that will also work. The sole reason I avoided that was because the calculations were messier, since we're integrating the square, and stuff. But apparently my example was not much better at reducing calculations. I also had to resort to WolframAlpha. $\endgroup$ – RandomStudent Mar 4 '19 at 19:23
  • $\begingroup$ Yeah the calculations are quite horrible to be honest. Even setting up the linear function for $f_n$ is a pain. $\endgroup$ – Sigurd Mar 4 '19 at 19:24
  • $\begingroup$ Ok, I was stupid. I have a way simpler example, let me post that as a second answer. $\endgroup$ – RandomStudent Mar 4 '19 at 19:26

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