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We are given the inner products defined as $\langle f, g\rangle _1 = \sum w_jf(x_j)g(x_j)$ and $\langle f, g \rangle _2 = \int_a^b w(x)f(x)g(x)dx$

Why is the inner product $\langle 1,x^k \rangle \neq 0$ for all $k \geq 2$ for the given inner products?

My attempt at a solution was using the fact that $\langle f, g + h \rangle = \langle f, g\rangle + \langle f, h\rangle$ so that $\langle 1, x^k \rangle = \langle 1, x^{k+0} \rangle$... but then I don't really know where to go from there. What am I missing?

Edit: $w(x)$ is a weighting function and $w$ is a set of weights. All weights are positive.

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    What is $w$ in this case? – 2017-02-24
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    What's true is that the inner product will be nonzero for even $k$; this is because $\langle 1, x^{2n} \rangle = \langle x^n, x^n \rangle$. But it can still be zero for odd $k$, for example if $a = -b$ and $w(x) = 1$. – 2017-02-24
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    It depends on your $w_i$ and $w(x)$ as well $a$ and $b$. – 2017-02-24
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    $w$ is a set of weights, all positive. Thanks! – 2017-02-24
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    $a$ and $b$ are both positive as well. – 2017-02-24
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    How do we show that the inner product is 0 for odd $k$ as well, given that $a$ and $b$ are both positive? – 2017-02-24
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    @snailshell remember that sum and integral are monotonous. So just compute some simple lower bounds. I guess $w(x)$ is continuous? – 2017-02-24

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