Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
You made a mistake when integrating by parts– you should have $$ \log (x) f(x)^2 - \int 2 f(x) \log (x)\,dx$$ – Omnomnomnom Oct 20 '13 at 17:05
up vote 3 down vote accepted

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?

share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.