Convolution operator has norm 1 with respect to $L^2$ norm

Question

Let $C_c(\mathbb{R})$ be the following:

$$C_c = \{ f \in C(\mathbb{R}) \mid \exists \text{ } T > 0 \text{ s.t. } f(t) = 0 \text{ for } |t| \geq T\}$$

Let $T_n \in L(C_c(\mathbb{R}))$ be a linear operator such that:

$$T_n u = \delta_n *u, \forall u \in C_c(\mathbb{R}),$$

where

$$ \delta_n(t)= \begin{cases} n^2(t+1/n) & -1/n \leq t \leq 0 \\ -n^2(t-1/n) & 0 < t \leq 1/n \\ 0 & \text{elsewhere} \end{cases} $$

I have to prove that with respect to the 2-norm, the operator $T_n$ has $\|T_n\| = 1$ and I have succeeded in proving that $\|T_nu\|_2 \leq \|u\|_2, \forall u \in C_c(\mathbb{R}).$

Now, I remained to prove that $\exists \text{ } u \in C_c(\mathbb{R}) \text{ s.t. } \|T_nu\|_2 = \|u\|_2$, but I really don't get how $u$ should be shaped in order to satisfy it.

Answer 1

Probable solution:

What we are interested in is proving that:

$$\lim_{m\to\infty} \|T_nu_m\|_2 = 1$$

If we write $u_m(t) = a_mv_m(t) $, where $v_m(t)$ defined as the hint by @mathcounterexamples.net, we can substitute in the limit (for simplicity we use $\|T_nu_m\|_2^2$):

$$\lim_{m\to\infty} \|T_nu_m\|_2^2 = \lim_{m\to\infty} {a_m}^2\|T_nv_m\|_2^2, $$

where ${a_m}^2 = \frac{1}{\|v_m\|_2^2} = \frac{3m}{3m^2+2}$

Then, according to what @mathcounterexamples.net pointed out, the integral can be split as:

$$\|T_nv_m\|_2^2 = \int_{-\infty}^{\infty}|(\delta_n*v_m)(t)|^2dt $$

$$ = \int_{-1/n-1/m}^{1/n}(\delta_n*v_m)^2(t)dt + \int_{1/n}^{m-1/n}(\delta_n*v_m)^2(t)dt + \int_{m-1/n}^{1/n+1/m+m}(\delta_n*v_m)^2(t)dt = \int_{-1/n-1/m}^{1/n}(\delta_n*v_m)^2(t)dt + \int_{1/n}^{m-1/n}dt + \int_{m-1/n}^{1/n+1/m+m}(\delta_n*v_m)^2(t)dt $$

Plugging it back in the limit:

$$\lim_{m\to\infty} {a_m}^2\bigg(\int_{-1/n-1/m}^{1/n}(\delta_n*v_m)^2(t)dt + \int_{1/n}^{m-1/n}dt + \int_{m-1/n}^{1/n+1/m+m}(\delta_n*v_m)^2(t)dt\bigg)$$

But in the first and in the third integral:

$$0 \leq (\delta_n*v_m)(t) \leq 1 \Rightarrow 0 \leq (\delta_n*v_m)^2(t) \leq 1$$

Therefore, for the first integral:

$$\lim_{m\to\infty} {a_m}^2\int_{-1/n-1/m}^{1/n}(\delta_n*v_m)^2(t)dt \leq \lim_{m\to\infty} {a_m}^2\bigg(\sup_{t \in [-1/n-1/m,1/n]}(\delta_n*v_m)^2(t)\bigg)^2(2/n+1/m) = \lim_{m\to\infty} {a_m}^2(2/n+1/m) = 0$$

And for the third integral as well:

$$\lim_{m\to\infty} {a_m}^2\int_{m-1/n}^{m+1/m+1/n}(\delta_n*v_m)^2(t)dt \leq \lim_{m\to\infty} {a_m}^2\bigg(\sup_{t \in [m-1/n,m+1/m+1/n]}(\delta_n*v_m)^2(t)\bigg)^2(2/n+1/m) = \lim_{m\to\infty} {a_m}^2(2/n+1/m) = 0$$

Finally we are left with:

$$\lim_{m\to\infty} {a_m}^2\bigg(\int_{1/n}^{m-1/n}dt\bigg) = \lim_{m\to\infty} {a_m}^2(m-2/n) = \lim_{m\to\infty}\frac{3m}{3m^2+2}(m-2/n) = 1 $$

Answer 2

Hint

Consider the function

$$v_m(x)=\begin{cases} 0 & x \le -\frac{1}{m}\\ 1+mx& -\frac{1}{m} \le x \le 0\\ 1 & 0\le x \le m\\ m^2-mx+1& m\le x \le m+\frac{1}{m}\\ 0 & m+\frac{1}{m} \end{cases}$$

And the function $u_m =a_m v_m $ with $a_m$ chosen such that $\Vert u_m \Vert_2=1$. You’ll be able to prove that

$$\lim\limits_{m \to \infty} \Vert T_n u_m\Vert_2 =1.$$

Notice in particular that: $$T_nv_m(x)= \begin{cases} 0 & x \le -\frac{1}{n} - \frac{1}{m}\\ 1 & \frac{1}{n} \le x \le m -\frac{1}{n}\\ 0 & m+\frac{1}{n} + \frac{1}{m} \le x \end{cases}$$ and $0 \le T_nv_m(x) \le 1$ elsewhere.

For example for $\frac{1}{n} \le x \le m -\frac{1}{n}$:

$$T_nv_m(x)=\int_{-\infty}^\infty \delta_n(x-t)v_m(t) \ dt = \int_{x-\frac{1}{n}}^{x+\frac{1}{n}} \delta_n(x-t)v_m(t) \ dt = \int_{x-\frac{1}{n}}^{x+\frac{1}{n}} \delta_n(x-t)\ dt=1$$