Showing that $(x^k)^\infty_{k=1}$ is a Cauchy sequence

Question

Let $V$ be the inner product space of polynomials in $x$, where $x\in[0,a]$ and $0<a<1$, with the standard inner product defined by an integral.

I need to show that the sequence $(x^k)^\infty_{k=1}$ is a Cauchy sequence, and that it converges to the zero function.

I know that for a Cauchy sequence, all the terms will eventually become close to each other, but I do not understand how this happens with the given sequence. Surely as $k$ approaches infinity, each $x^k$ will be infinitely bigger than the last?

Answer 1

The inner product, $\langle f,g\rangle$ for two polynomials $f(x)$ and $g(x)$ is given by the integral $\langle f,g\rangle = \int_0^a f(x)g(x)\,dx$. For inner product spaces, the distance between two points is given by $|f-g|^2 = \langle(f-g),(f-g)\rangle$. It follows then, that the distance between $x^m$ and $0$ is ???.

So if we show that is converges to the zero function, what can we say about the "cauchyness" of the sequence?

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Edit: There was some ambiguity, so for two functions the inner product, $f\cdot g = \langle f,g\rangle$ is given $\int_0^a f(x)g(x)\,dx$, not the normal product. I have edited my post to reflect this

Answer 2

Suppose $a \in (0,1)$.

Let $\varepsilon>0$ be given.

Since $\{x^k\}_{k=1}^\infty$ converges uniformly to $0$ on $[0, a]$ (a compact subset of $[0,1)$), the sequence is uniformly Cauchy on $[0, a]$ (every convergent sequence is Cauchy). In other words, we may find $N \in \mathbb{N}$ so that

$$\sup_{x\in [0,a]} \left|x^m-x^n \right|<\frac{\varepsilon}{\sqrt{a}} \text{ whenever } m,n \geq N.$$

So if $m,n \geq N$, then we have $$\int_0^a \left|x^m-x^n \right|^2 dx \leq \frac{\varepsilon^2}{a} \int_0^a dx=\varepsilon^2.$$

We may take the square root on both sides of the above inequality to have $$\left[ \int_0^a \left|x^m-x^n \right|^2 dx \right]^{1/2} \leq \varepsilon\text{ whenever } m,n \geq N.$$