求复合算子$T=A(id-B): L^2[0,1] \to L^2[0,1]$的算子范数

API 探路者

2026-5-19

Nice work getting a valid upper bound of 2—let’s tighten that up and find the exact operator norm of $T$, which turns out to be 1. Here's how to get there:

First, let's restate the operator clearly for reference:
$$Tf(x) = A(\text{id} - B)f(x) = x\left(f(x) - \int_0^1 f(t)dt\right) = xf(x) - x\int_0^1 f(t)dt$$

Step 1: Prove $|T| \leq 1$

Take any $f \in L^2[0,1]$, and let $c = \int_0^1 f(t)dt$ (the average value of $f$). We start by computing the squared norm of $Tf$:
$$|Tf|_2^2 = \int_0^1 x^2 |f(x) - c|^2 dx$$

Since $x^2 \leq 1$ for all $x \in [0,1]$, we can immediately bound this:
$$|Tf|_2^2 \leq \int_0^1 |f(x) - c|^2 dx$$

Now expand the right-hand side using the definition of $c$ and properties of inner products:
$$\int_0^1 |f(x) - c|^2 dx = |f|_2^2 - 2\operatorname{Re}\left(c \cdot \overline{\int_0^1 f(t)dt}\right) + |c|^2 = |f|_2^2 - |c|^2$$

Because $|c|^2 \geq 0$, this simplifies to:
$$|Tf|_2^2 \leq |f|_2^2 \implies |Tf|_2 \leq |f|_2$$

This tells us the operator norm $|T|$ is at most 1.

Step 2: Show $|T| \geq 1$ (using a sequence of test functions)

To confirm the norm is exactly 1, we need a sequence of functions ${f_k}$ where $|f_k|_2 = 1$ and $|Tf_k|_2$ approaches 1 as $k$ grows.

Define the normalized functions:
$$f_k(x) = \sqrt{2k+1} \cdot x^k$$

We can check this is normalized:
$$|f_k|_2^2 = (2k+1)\int_0^1 x^{2k}dx = (2k+1) \cdot \frac{1}{2k+1} = 1$$

Now compute $c_k = \int_0^1 f_k(t)dt = \sqrt{2k+1} \cdot \frac{1}{k+1}$, then calculate $|Tf_k|_2^2$:
$$|Tf_k|_2^2 = \int_0^1 x^2 |f_k(x) - c_k|^2 dx$$

Expanding this out and substituting the integrals we need:
$$
\begin{align*}
|Tf_k|_2^2 &= (2k+1)\int_0^1 x^{2k+2}dx - 2c_k\sqrt{2k+1}\int_0^1 x^{k+2}dx + |c_k|^2\int_01 x^2 dx \
&= \frac{2k+1}{2k+3} - \frac{2(2k+1)}{(k+1)(k+3)} + \frac{2k+1}{3(k+1)^2}
\end{align*}
$$

As $k \to \infty$:

The first term $\frac{2k+1}{2k+3}$ approaches 1
The second and third terms decay to 0 (they're proportional to $\frac{1}{k}$ and $\frac{1}{k^2}$ respectively)

So $|Tf_k|_2^2 \to 1$, which means $|Tf_k|_2 \to 1$.