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Show $$\langle f,g \rangle := \frac 1 {2\pi} \int^{\pi}_{-\pi} f(x) \overline {g(x)} dx$$

define an inner product on the complex vector space of continuous functions.

I must establish $\langle f,g \rangle = \overline {\langle g,f \rangle}$, however I get to the point $$\langle g,f \rangle := \frac 1 {2\pi} \int^{\pi}_{-\pi} g(x) \overline {f(x)} dx =\frac 1 {2\pi} \int^{\pi}_{-\pi} g(x) \overline {f(x)} dx = \frac 1 {2\pi} \int^{\pi}_{-\pi} \overline {f(x)} g(x) dx=\int^{\pi}_{-\pi} \overline {f(x) \overline {g(x)}} dx$$, and I want to take the conjugate "out" the integral, but how ?

Also, how can I verify that $\langle f,f\rangle = 0 \iff f = 0$ ?

Shuzheng
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  • Good question, I remember dealing with this as well. I don't remember the exact answer I gave, but I think it was not rigorous. – Ruben May 07 '14 at 07:25
  • For the conjugate, I think you can just do: $ f(z) = u(z) + iv(z) $ where $u$ and $v$ are $\mathbb{C} \rightarrow \mathbb{R}$, so $\int{f(z)dz} = \int{u(z)dz} + i\int{v(z)dz}$ and that justifies that you can take the conjugate out of the integral. – Ruben May 07 '14 at 07:27
  • But I found this link: http://math.stackexchange.com/questions/197642/does-the-complex-conjugate-of-an-integral-equal-the-integral-of-the-conjugate - People says it cannot be done ? – Shuzheng May 07 '14 at 07:30
  • And for the second one, you have (by continuity of f) that if there is a $z_0$ such that $f(z_0) \not= 0$, there is a neighbourhood of $z_0$ where $f$ is nonzero. $f$ multiplied by it's conjugate is always zero, and I think you can show that the integral must be more than (size of the neighbourhood) * (lowest nonzero value of f * in the neighboorhood). – Ruben May 07 '14 at 07:32
  • In your case, $x$ is a real variable, so $\overline{\mathrm dx}=\mathrm dx$. So you don't need to conjugate the differential. – Christoph May 07 '14 at 07:32
  • Ahh, so the link I've posted are concerned with a complex variable integral ? – Shuzheng May 07 '14 at 12:21

3 Answers3

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It's ok to "take the conjugate out of the integral". After all, if $f(x) = u(x)+iv(x)$, where $u$ and $v$ are real-valued then

\begin{align} \int_a^b \overline{f(x)}\,dx &= \int_a^b \big( u(x)-iv(x) \big)\,dx \\ &= \int_a^b u(x)\,dx - i\int_a^b v(x)\,dx \\ &= \overline{\int_a^b f(x)\,dx}. \end{align}

Secondly if $\langle f,f \rangle = 0$, then $$ \int_{-\pi}^\pi |f(x)|^2\,dx = 0. $$ For a slick way to see that this implies $f=0$, define $$ g(t) = \int_{-\pi}^t |f(x)|^2\,dx. $$ Clearly, $g(-\pi) = g(\pi) = 0$. On the other hand, the integrand is positive, so $g$ is increasing. Hence $g(t) = 0$ for all $t$, and conequently $g'(t) = 0$ for all $t$. On the other hand, the fundamental theorem of calculus implies that $g'(t) = |f(t)|^2$, so $|f(t)| = 0$ for all $t$, which in turn implies that $f(t) = 0$ for all t.

mrf
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  • Maybe an off-topic, but if the complex integral of $f(x)$ exist then $u(x)$ and $v(x)$ must be continuous. Thus the integral of $\overline {f(x)}$ exist. But how does one justify the complex integral of $f(x)\overline {g(x)}$ does exist ? Here we get a product of continuous functions ? – Shuzheng May 07 '14 at 07:38
  • Well, $u$ and $v$ don't have to continuous in order for the complex integral to exist (but the converse is certainly true). So, if $f$ and $g$ are continuous, then $f\bar g$ is also continuous, and there is no problem with the existence of the integral. – mrf May 07 '14 at 07:40
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For any function $f$:

$$\int \overline{f(x)} dx = \overline{\int f(x) dx}$$

To see this, write $f(x) = g(x) + ih(x)$ with $g$ and $g$ being real-valued functions.

$$\int g(x) - ih(x) dx = \int g(x) dx - i \int h(x) dx$$

Now, to prove that $\langle f,f \rangle = 0 \implies f = 0$, notice that $\langle f,f \rangle$ is:

$$\frac{1}{2\pi} \int_{-\pi}^\pi f(x) \overline{f(x)} dx = \frac{1}{2\pi} \int_{-\pi}^\pi g(x)^2 + h(x)^2 dx$$

$g(x)^2 + h(x)^2$ is always positive, but the integral evaluates to $0$. The rule for continuous functions is that if a nonnegative continuous function has integral $0$, then it is $0$ everywhere. So $g(x)^2 + h(x)^2 = 0$. I think you can fill in the rest of the details.

A.S
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For the first question, note that $x$ is a real variable, so your complex integral is just defined as $$ \int_{-\pi}^\pi f(x)\,\mathrm dx = \int_{-\pi}^\pi \operatorname{Re}(f(x))\,\mathrm dx + i\int_{-\pi}^\pi \operatorname{Im}(f(x))\,\mathrm dx. $$ Thus conjugating $f$ under the integral is the same as conjugating the whole integral.

Christoph
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