Determine whether $\langle f, g\rangle = \int_1^e {1 \over x} f(x)g(x)\,dx $ is a inner product.

Question

Let $C[1,e]$ be the set of continuous real-valued functions with domain $W:=[1,e]$.

Let $$\langle f, g\rangle = \int_1^e {1 \over x} f(x)g(x)\,dx $$ be a function.

Determine whether $\langle \cdot, \cdot\rangle$ is an inner product. (i.e. $(a,b): W \times W \to C$ (C is complex set)

In order that it be a inner product it has to meet 3 conditions:

1. $\forall f \in W $ , an inner product of $(f,f) \ge 0 $, $f = 0 \iff (f,f) = 0$

So I tried to solve this integral:

$$\int {1 \over x}f(x)f(x)\,dx = \int {1 \over x}f(x)^2\,dx = {1 \over x}f(x)^2 - \int{1 \over x}2f(x) = {f(x)^2 \over x} - 2 \int {f(x) \over x} $$

Then I got stuck. can you please help? thanks in advance.

Answer 1

First, this is a good question! You said what step you're stuck at, and didn't try to get help solving the entire problem before you get through that step. I wish everybody here did that. Kudos.

It's not the right approach to try to calculate the integral. You can't—$f(x)$ is arbitrary, and could be something quite terrible and complicated (and check your steps!—they don't work). But let's look again at what we'd like to prove:

$$ \int_1^e \frac{1}{x} f(x)^2 dx \geq 0$$

Again, don't start manipulating the equation until you have a plan. How would we show this? How do you tell, just from looking at it, that an integral can't be negative?

Pay attention to the domain of integration—if you change $1$ and $e$ to something else, the problem might be false. Try plugging in some really simple functions if you need to (like $f(x)=x$). Maybe draw a picture.

But there aren't any complex mathematics needed here. Just one simple realization.

Think about what an integral is. Can integrals ever be negative? How? Think about the reverse question: What conditions should $g(x)$ meet to guarantee that $\int_a^b g(x)dx$ is always non-negative?

Answer 2

Don't think in terms of evaluating the integrals; think in terms of using the properties of integrals to your advantage.

For example: we can show that first condition holds based on the following fact:

• The integral of a non-negative continuous function over an interval is zero if and only if that function is (identically) zero everywhere on that interval

Now, we can make the following argument: suppose $f(x)$ is non-zero function over $[1,e]$. Then $f(x)^2$ is a non-negative function that is not identically $0$. Thus, $\frac 1x f(x)$ is a non-negative function that is not identically zero, which means that $\int_1^e \frac 1x f(x)^2 dx$ is non-zero.

Now, suppose $f(x) = 0$. Then $\int_1^e \frac 1x f(x)^2 dx = 0$.

Thus, the first property is satisfied.