
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

Re: integrals squared
Hi hannah,
We can get what you want by applying the CauchySchwarz inequality (see Cauchy). The wikipedia article gives the statement of the theorem in terms of arbitrary innerproduct 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.
Realvalued continuous functions defined on [0,1] form a vector space over the real numbers (meaning if we add two realvalued continuous functions defined on [0,1] we again get a continuous realvalued function on [0,1], and if we multiply a realvalued continuous function defined on [0,1] by a real scalar we get another realvalued 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 innerproduct to the vector space. The innerproduct 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 innerproduct 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!

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!