# Questions about operator norm on [0,1]

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!

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