# positive definiteness of an inner product

I am having trouble understanding the solution for an excercise I did in my linear algebra class.

The question is: $$\text{Let }V \text{ be the vectorspace of the real polynominals of degree} \leq n. \\ \text{Show that } \langle \textit{p},\textit{q}\rangle := \int_{0}^{\infty} p(t)q(t)e^{-t} dt \text{ defines an inner product on } V$$

Symmetry and linearity follows easily. For the positive definiteness i try to show that: $$\langle \textit{p},\textit{p} \rangle = \int_{0}^{\infty} p(t)^2e^{-t} dt \geq 0 \text{ with equality only if } p=0$$ I can see easily that $$e^{-t} > 0$$ is true but i have trouble showing that $$p(t)^2 >0$$

The solution is as follows: $$\text{Let } p \in V\backslash\{0\} \text{. Choose a point } x_0 \text{ such that } p(x_0)\neq 0. \text{ Since }p \text{ induces a continous function it holds that:}\\ \vert p(x) \vert \geq c := \frac{1}{2} \vert p(x_0) \vert > 0 \text{ on an interval } [a,b] \subset \mathbb{R}^{\geq0} \text{ with } x_0 \in [a,b] \text{ and therefore: } \\ \langle p, p \rangle = \int_{0}^{\infty} p(t)^2e^{-t} dt \geq \int_{a}^{b} p(t)^2e^{-t} dt \geq c \cdot e^{-b} \cdot (b-a) > 0$$

If someone can explain to me, how you can get to those inequalities from the fact that $$p(t)^2$$ is continous, it would be greatly appreciated.

• A polynomial is always continuous. In general, for a continuous function $f:\mathbb{R}\rightarrow\mathbb{R}$, if $f(x_0)>0$, then there exists $\varepsilon>0$ such that $f(x_0)-\varepsilon >0$. Then $f^{-1}(]f(x_0)-\varepsilon,f(x_0)+\varepsilon[)$ is open since $f$ is continuous. Mar 29, 2022 at 11:45

For a continuous function the value $$p(x)$$ has to be "close" to the value $$p(x_0)$$ if $$x$$ is "close" to $$x_0$$. More precisely: for each $$\varepsilon >0$$ there is a $$\delta>0$$ such that $$|p(x)-p(x_0)| < \varepsilon$$ for all $$x$$ s.t. $$|x-x_0| < \delta$$. That is the definition of continuity in the point $$x_0$$.
Choosing $$\varepsilon = \frac 12 |p(x_0)|$$ (where $$x_0$$ is a point with $$p(x_0) \ne 0$$) we get a small interval (determined by the $$\delta$$ corresponding to our $$\varepsilon$$) in which $$|p(x)-p(x_0)| < \frac 12 |p(x_0)|$$ which implies that $$|p(x)| \ge \frac 12 |p(x_0)|$$ on this interval - let's call it $$(a,b)$$. Then we can proceed as in the solution given by you: $$\langle p,p \rangle = \int_0^\infty p(t)^2 e^{-t} dt \ge \int_a^b p(t)^2e^{-t} \ge \frac 14 |p(x_0)|^2 e^{-b} \cdot (b-a) >0$$ because $$p(x_0) \ne 0, e^{-b} >0$$ and $$b>a$$.
This gives the positive definiteness because it says that any $$p \ne 0$$ will have $$\langle p,p \rangle >0$$.
• Is there a reason that $\frac{1}{2}$ has been chosen as a scalar or is it "random" ? Mar 29, 2022 at 11:48
• It is "random" - the $\varepsilon$ can be choosen freely. But for this purpose the choice of $\frac 12 |p(x_0)|$ is a "good" choice because it gives us an interval where $p$ is still $\ne 0$. If we chose for example $\varepsilon = 2 |p(x_0)|$ that would only give us an interval in which $-p(x_0) \le p(x) \le 3 p(x_0)$ so $p(x)$ could be $=0$. For such an argument it is convenient to take half of $|p(x_0)|$, but I could have chosen any number $a < |p(x_0|$ for the argument to work. Mar 29, 2022 at 11:51