I am struggling with the following question:

Consider the space $V$ of continuous functions on [0,1] with the 2-norm $‖f‖_2^2$=$∫_0^1|f|^2$. $V$ is an incomplete normed linear space. For a continuous function φ on [0,1], define a linear map $M_φ:V⟶V$ by $M_φ f=φf$. Show that $M_\varphi$ is bounded and find its norm.

Here is what I did to show $M_\varphi$ is bounded.

$‖φf‖_2^2$=$∫_0^1|φf|^2$$‖φ‖_∞^2$ $∫_0^1|f|^2$ =$‖φ‖_∞^2$ $‖f‖_2^2$

This shows that $‖M_φ ‖≤‖φ‖_∞$, hence $M_φ$ is bounded.

I think in order to find the operator norm, it must be shown that the opposite inequality, $‖M_φ ‖\ge‖φ‖_∞$ hold. I suspect the unit vector in $C([0,1])$ must somehow be used. Here is my rough outline:

  1. Let $g$ be a unit vector in $C([0,1])$
  2. Fix $ε$>0. Find $x_0\in [0,1]$ such that $g(x_0)=1$ and $‖φ‖_∞-ε≤|φ(x_0 )|≤‖φ‖_∞$

  3. Show that $ ‖M_φ g‖_\infty \ge ‖φ‖_∞ -ε$

  4. Since $ ‖M_φ g‖_2 \ge ‖M_φ g‖_\infty$, we have $‖M_φ ‖≥‖φ‖_∞$

I know there is something amiss, could you please help me understand where went wrong? Thank you so much in advance, have a wonderful day!

  • $\begingroup$ what is $M_\varphi$?? And what is $\varphi$? $\endgroup$ – Masacroso Jun 10 '19 at 21:01
  • $\begingroup$ Sorry I forgot to include, will edit now $\endgroup$ – mathnoob777 Jun 10 '19 at 21:02

Since $\lvert \varphi \rvert$ is continuous, it achieves its maximum somewhere on $[0,1]$; say the maximum occurs at $x_0$ so that $\|\varphi\|_\infty = \lvert \varphi(x_0)\rvert $. Define $f_n:[0,1]\to [0,\infty)$ by $$f_n(x)^2 = \left\{ \begin{matrix}0, & 0 \le x < x_0 - \tfrac 1 n,\\ n^2(x-x_0 +\tfrac 1 n), & x_0 - \tfrac 1 n \le x < x_0, \\ n^2(x_0 +\tfrac 1 n - x), &x_0 \le x < x_0 + \tfrac 1 n,\\ 0, &x_0+\tfrac 1n \le x. \end{matrix}\right.$$ Each $f_n$ is continuous and the graph of $f_n^2$ is a triangular spike centered at $x_0$ with base $2/n$ and height $n$, and thus $\int^1_0 f_n(x)^2 dx = 1$.

Fix $\epsilon > 0$. Since $\varphi$ is continuous, there is $\delta >0$ such that when $\lvert x - x_0 \rvert < \delta$, we have $\lvert \varphi(x)\rvert \ge \|\varphi\|_\infty - \epsilon$. Choose $n$ large enough that $\frac 1 n < \delta$. Then \begin{align*}\|M_\varphi f_n\|^2_{2} &= \int^1_0 \lvert f_n(x) \rvert^2 \lvert \varphi(x) \rvert^2dx \\ &= \int_{x_0 - \tfrac 1n}^{x_0+\tfrac 1n}\lvert f_n(x) \rvert^2 \lvert \varphi(x) \rvert^2dx \\ &\ge (\|\varphi\|_\infty - \epsilon)^2\int_{x_0 - \tfrac 1n}^{x_0+\tfrac 1n}\lvert f_n(x) \rvert^2dx = (\|\varphi\|_\infty-\epsilon)^2. \end{align*} Taking the square root shows that $\|M_\varphi\| \ge \|M_{\varphi}f_n\|_2 \ge \|\varphi\|_\infty-\epsilon$. Since $\epsilon >0$ was arbitrary, send it to zero and you have your result.

EDIT: Obvious modifications can be made in the cases that $x_0 = 0$ or $x_0 =1$. Also, there is no reason you need so explicit a function $f_n$; any function $f$ which is positive for $\lvert x-x_0\rvert < \delta$ and zero when $\lvert x-x_0 \rvert > \delta$ should work.

  • $\begingroup$ I think I get it now, thank you so much for being so thorough, Sir/Madam!!! $\endgroup$ – mathnoob777 Jun 10 '19 at 21:37
  • $\begingroup$ Glad I could help! $\endgroup$ – User8128 Jun 10 '19 at 21:40

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