# Polynomial Basis with zero integral

I have a question and please I need help. First all, this is the context:

Let $p : T\rightarrow\mathbb{R}$ a polynomial of degree $\leq k$ with null measure condition onto T, that is $$\int_T p\ =\ 0,$$ where $T$ is some region in $\mathbb{R}^n$. I want to construct a basis of the subespacio where $p$ lies, namely $Q_{k,0}$.

Now suppose the I have $\{\varphi_0,\varphi_1,\ldots,\varphi_N\}$ an arbitrary basis for $P_k(T) =$ polynomial space of degree $\leq k$ onto $T$. Then, in order to obtain a basis for $Q_{k,0}$, I defined $$\psi_i\ =\ \varphi_i - \frac{1}{|T|}\int_T\varphi_i,\quad i = 0,1,\ldots,N.$$ Observe that $\{\psi_0,\psi_1,\ldots,\psi_N\}\subseteq Q_{k,0}$, but is not a basis, because they are linearly dependent.

So, my question is: how I can get a basis of $Q_{h,0}$ from the set $\{\psi_i\}$? In other words, how I know WHICH $\psi_i$ must be delete, to get a basis?

Remember, $\{\varphi_i\}$ is an arbitrary basis, because if we now that one $\varphi_i$ is constant, then the respective $\psi_i$ must be delete. The problem is, for example, for the Lagrange basis.

Thanks so much in advance for any hints.

Assuming that $\varphi_0 = c \neq 0$, i.e. that $\deg \varphi_0 = 0$, for your basis $(\varphi_i)$, the $\psi_i$ are linearly independent of you leave out $\psi_0$. By construction of the $\psi_i$, you have $$\psi_i = \varphi_i + \lambda_i\varphi_0 \quad\text{where } \lambda_i = -\frac{1}{c|T|}\int_T \varphi_i(t) \,dt$$ Assume that for some linear combination of the $\psi_i$, i.e. for some $\mu_1,\ldots,\mu_N$, you have $$\sum_{k=1}^N \mu_i \psi_i = \sum_{k=1}^N \mu_i\varphi_i + \varphi_0\underbrace{\sum_{k=1}^N \mu_i \lambda_i}_{=:\mu_0} = 0 \text{.}$$ Since the right-hand side is a linear combination of the $\varphi_i$ with coefficients $\mu_0,\mu_1,\ldots,\mu_N$, this implies $\mu_i = 0$ for all $i$, which shows that the $(\psi_i)_{1 \leq i \leq N}$ are linearly independent.
Thus the $(\psi_i)_{1 \leq i \leq N}$ generate a least a subspace of $Q_{k,0}$. In particular, for $N=k$, the generate a $k$-dimensional subspace. Now you just need to argue that $\dim Q_{k,0} = k$. This follows from the fact that the integral of a constant polynomial $p$ over $T$ is only zero if $p$ is identically zero, i.e. the zero polynomial.
In the general case, i.e. if nothing is known about the $\varphi_i$ except that they form a basis, you always can select $k$ linearly independent $\psi_i$, but I don't think you can describe which those are except by, well, saying "pick them linearly independent".
• Thanks for your answer. But, is a easy way to do the case deg $\varphi_i \neq i$ ? – FASCH Apr 28 '14 at 17:48
• @FASCH I've updated my answer to say $\varphi_0 = c$ - that's in fact the only one of the degree constraints that the rest of the answer used. Without any such constraint, I don't think there's a better characterization of the $\psi_i$ you have to pick except saying "Pick them linearly independent". For a specific basis $\varphi_i$, however, better characterizations might exist - but to explore that you'd first have to state the basis you have in mind. If you don't have a particular basis in mind, then why not just use $\varphi_i = x^i$? – fgp Apr 28 '14 at 18:56