2.14

复值函数内积

Stanford 工科课程 The Fourier Transforms and its Applications 的 Course Reader 的一个附录给出了对复值函数内积形式的说明，解释了为什么有共轭出现，内容如下：

Here’s an argument why the conjugate comes in defining a complex inner product. Let’s go right to the case of integrals. What if we apply the Pythagorean Theorem to deduce the condition for perpendicularity in the complex case, just as we did in the real case? We have

$${\int }_0^1|f(t)+g(t){|}^2={\int }_0^1|f(t){|}^{2}dt+{\int }_0^1|g(t){|}^{2}dt$$

$${\int }_0^1|f(t){|}^{2}dt+2\text{Re}({\int }_0^{1}f(t)\overline{g(t)}dt)+{\int }_0^1|g(t){|}^{2}dt={\int }_0^1|f(t){|}^{2}dt+{\int }_0^1|g(t){|}^{2}dt$$

So it looks like the condition should be

$$\text{Re}({\int }_0^{1}f(t)\overline{g(t)}dt)=0.$$

Why doesn’t this determine the definition of the inner product of two complex functions? That is, why don’t we define

$$(f,g)=\text{Re}({\int }_0^{1}f(t)\overline{g(t)}dt)?$$

This definition has a nicer symmetry property, for example, than the definition we used earlier. Here we have

$$(f,g)=\text{Re}({\int }_0^{1}f(t)\overline{g(t)}dt)=\text{Re}({\int }_0^1\overline{f(t)}g(t)dt)=(g,f),$$

so none of that Hermitian symmetry that we always have to remember.

The problem is that this definition doesn’t give any kind of homogeneity when multiplying by a complex scalar. If $\alpha$ is a complex number then

$$(\alpha f,g)=\text{Re}({\int }_0^1\alpha f(t)\overline{g(t)}dt)=\text{Re}(\alpha {\int }_0^{1}f(t)\overline{g(t)}dt).$$

But we can’t pull the $\alpha$ out of taking the real part unless it’s real to begin with. If $\alpha$ is not real then

$$(\alpha f,g)\ne \alpha (f,g).$$

Not having equality here is too much to sacrifice. (Nor do we have anything good for $(f,\alpha g)$, despite the natural symmetry $(f,g)=(g,f)$.) We adopt the definition

$$(f,g)={\int }_0^{1}f(t)\overline{g(t)}dt.$$

另外，编辑这段内容时，发现 obsidian 框选段落变成引用格式，会在 $LaTeX$ 公式内部的行首也添加 ">"，导致格式错误

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