# integrals squared

• Dec 17th 2012, 05:16 AM
hannahcooper
integrals squared
I'm working through a proof for part of my university project and I'm a little stuck understanding part of it:
if:
$\displaystyle \int_0^1 (f(x))^2 dx =\left(\int_0^1 f(x)dx \right)^2$
then f(x) = constant
why does this prove f(x) is constant?
Thankyou
• Dec 17th 2012, 11:01 AM
GJA
Re: integrals squared
Hi hannah,

We can get what you want by applying the Cauchy-Schwarz inequality (see Cauchy). The wikipedia article gives the statement of the theorem in terms of arbitrary inner-product spaces; if you know what these are and how they apply to what you've asked about, then read no further. If not I have provided an explanation below about how it helps us establish the result.

Real-valued continuous functions defined on [0,1] form a vector space over the real numbers (meaning if we add two real-valued continuous functions defined on [0,1] we again get a continuous real-valued function on [0,1], and if we multiply a real-valued continuous function defined on [0,1] by a real scalar we get another real-valued continuous function defined on [0,1]); this vector space is typically denoted by $\displaystyle C([0,1], \mathbb{R})$. When we have a vector space over a field it is often desirable to attach what is known as an inner-product to the vector space. The inner-product allows us to define geometric quantities on our space like lengths of vectors, angles between vectors, etc. The "dot product" you learned about in multivariable Calculus is an inner-product on $\displaystyle \mathbb{R}^{3},$ for example. The (standard) inner product for $\displaystyle C([0,1], \mathbb{R})$ is given by

$\displaystyle <f,g>=\int_{0}^{1}f(x)g(x)dx,$

where $\displaystyle f(x),g(x)\in C([0,1],\mathbb{R}).$

Now, as to how all of this pertains to your question, if we take $\displaystyle g(x)=1$ to be the constant function 1, the Cauchy Schwarz inequality says that

$\displaystyle \left(\int_{0}^{1}f(x)\cdot 1 dx \right)^{2}\leq \int_{0}^{1}f(x)^{2}dx\cdot\int_{0}^{1}1^{2}dx= \int_{0}^{1} f(x)^{2}dx,$

where equality holds above if and only if $\displaystyle f(x)$ and $\displaystyle g(x)=1$ are linearly dependent. According to the assumption you've provided we do have equality above. Hence, $\displaystyle f(x)$ and $\displaystyle g(x)=1$ are linearly dependent; i.e. $\displaystyle f(x)=c\cdot g(x)=c\cdot 1=c.$ We see that $\displaystyle f(x)=c$, so $\displaystyle f$ is constant.

Does this answer your question? Let me know if anything is unclear. Good luck!
• Dec 18th 2012, 02:17 AM
hannahcooper
Re: integrals squared
Thankyou that is massively helpful, it was the linear dependence bit I was missing, so all makes a lot more sense now!