1
$\begingroup$

Can anyone help me prove that the $L^2$ inner product is in fact an inner product?

I'm particularly struggling to prove that it is conjugate-symmetric and that length is positive.

This is the inner product in question:

For $f,g \in L^2([a,b])$, $\langle f,g \rangle = \int_a^b f(t) \overline{g(t)}dt$

  • 1
    For positive length, just stare at $\langle f,f \rangle$ for some time. Squares of functions always take positive values. – 2017-01-31
  • 1
    But how to do you show that if $\langle f , f \rangle = 0$ then $f=0$, I am still stuck here. – 2017-01-31

1 Answers 1

4

Note that $\langle f,f\rangle = \int_a^b f(t) \overline{f(t)} dt = \int_a^b |f(t)|^2 dt$. If $f \neq 0$, then $f^2 \geq 0$ and is strictly greater than zero in some non-zero measure set, so $\int_a^b f(t) \overline{f(t)} dt > 0$. On the other hand, suppose that $\int_a^b |f(t)|^2 dt = 0$. Since $|f(t)|^2 \geq 0$ for all $t$, it follows that $f(t) = 0$ almost everywhere, because if it is not zero for some non-zero measure set, then the integral will become greater than zero, as we explained earlier.

EDIT : To clarify (love your name!)KingDingeling's doubt, albeit belatedly.

The thing is, you have got to understand what equality means in $L^2([a,b])$ (or in $L^p$ spaces, for that matter). What does it mean for $f = g$ in $L^2([a,b])$? Does it mean that $f(x) = g(x)$ for all $x \in [a,b]$?

NO. What it means, is that the set $\{x : f(x) \neq g(x)\}$ has measure zero. (The measure is the one with respect to which the integral $\mbox{dt}$ is defined. It is not mentioned in the question, we will calll it $\mu$).

For example, let $[a,b] = [0,1]$, and let us work wih the Lebesgue measure. Let us define two functions, $f(x) \equiv 0$ (identically zero) and another function $g(x) = 0$ if $x \neq 0$ and $g(0) = 1$. Clearly, $f(0) \neq g(0)$, so they are not equal in the usual sense. However, the set in which they differ is $\{0\}$, and the Lebesgue measure of $\{0\}$ is zero. Therefore, $f=g$ as elements of $L^2([a,b])$.

Therefore, while we integrate functions, the elements of $L^2([a,b])$ are in fact equivalence classes under the relation of being equal except on a measure zero set!

Now, let us understand what is going on.

"If $f \neq 0$" means "as elements of $L^2$". So now, the fact that $f \neq 0$, means that there is a set $A$ with positive measure such that $f(x) \neq 0$ for all $x \in A$.

The fact that $|f^2| \geq 0$ is ok to see. However, on the set $A$ we also have $|f(x)|^2 > 0$, because $f(x) \neq 0$.

From this, we want to conclude that $\int_A |f(x)|^2dx > 0$. How do we do this? From monotonicity one may argue $\geq 0$. But we need a standard trick at this point.

The trick is nice : let $A_n = A \cap \{|f(x)|^2 \geq \frac 1n\}$, $n \in \mathbb N$. Note that $\cup_{i=1}^{\infty} A_i = A$, because if $x \in A$ then $|f(x)|^2 > 0$ so $|f(x)|^2 \geq \frac 1n$ for some $n$.

Now, if all the $A_n$ have measure zero, then $A$ being the increasing limit of the $A_n$ will have measure zero as well (property of a measure), which cannot happen. Consequently, some $A_N$ has positive measure, say $\mu(A_N) = \epsilon > 0$.

Now, on $A_N$, there is domination of $|f|^2$ over $\frac 1N$, therefore : $$ \int_{A_N} |f(x)|^2dx \geq \int_{A_N} \frac 1N dx = \frac{\mu(A_N)}{N} = \color{orange}{\frac{\epsilon}{N} >0} $$

Now on $A_N^c$ w anyway have domination of zero, therefore : $$ \int_a^b |f(t)|^2 dt = \int_{A_N} |f(t)|^2 dt + \int_{A_N^c} |f(t)|^2 dt \geq \frac \epsilon N + 0 = \frac \epsilon N > 0 $$

Therefore, the inner product $\langle f,f\rangle > 0$.

The rest is clear.