# Show that $U^\perp=\{0\}$.

Background

Hey everyone. I'm absolutely stumped on an exercise I am working on out of Axler's Linear Algebra Done Right, 3rd edition. Funnily enough, the sample chapter available on his website is the one I am currently working in.

The Problem

If are you following along from the sample chapter, the problem in question is Exercise 14(a) of Section 6.C. It asks the following:

Suppose $C_{\mathbb{R}}([-1,1])$ is the vector space of continuous real-valued functions on the interval $[-1,1]$ with inner product given by $$\langle f,g \rangle=\int_{-1}^1f(x)g(x)\,dx$$ for $f,g\in C_{\mathbb{R}}([-1,1])$. Let $U$ be the subspace of $C_{\mathbb{R}}([-1,1])$ defined by $$U=\left\lbrace f\in C_{\mathbb{R}}([-1,1]) \ | \ f(0)=0 \right\rbrace.$$ Show that $U^\perp=\{0\}$.

What I Know

I understand that this is an infinite-dimensional vector space (right?) and thus the theorems provided in the accompanying chapter do not apply, since those are made with the assumption that the vector space you are working with is finite. Thus, you need to find a way to show this without using those theorems.

My Confusion

I've looked all over this site and the web for some potential hints as to how you go about doing this. While I couldn't find my exact question, I did find some related ones. Many of these similar questions seemed to utilize the fact that for a continuous real-valued vector space $V$ you can write $$V=U_{even}\oplus U_{odd}$$ where $U_{even}$ is the subspace of $V$ consisting of all even functions and $U_{odd}$ is the subspace of $V$ consisting of all odd functions. Please correct me on that if I am wrong. This seems potentially helpful, but I'm not sure how to use it if it is. One other solution seemed to utilize methods that I somewhat recall from real analysis. However, I'd like to do this proof using the methods of linear algebra, if that is at all possible. I think if I understood how to characterize $U$ better, then I might understand how to go about this.

How would you start this problem?

• What you have to do is this: for each nonzero continuous function $f$ on $[0,1]$ find a continuous function $g$ on $[0,1]$ with the properties that $g(0)=0$ and $\int_0^1 f(x)g(x)\,dx\ne0$. This is an analysis question. Apr 15 '18 at 6:10

This is not really a linear algebra problem. It is an analysis problem that involves definitions from linear algebra. So in order to solve it, you're mainly going to have to do analysis; the linear algebra is just a matter of writing down what the definitions mean.

So, let's write down the definition of $U^\perp$. By definition, it's the set of all $g\in C_\mathbb{R}([-1,1])$ such that $\langle f,g\rangle =0$ for all $f\in U$. To prove $U^\perp=\{0\}$, you want to prove that $0$ is the only element of $U^\perp$. In other words, you want to prove that if $g\neq 0$, then $g\not\in U^\perp$. To prove $g\not\in U^\perp$, you want to prove there exists $f\in U$ such that $\langle f,g\rangle\neq 0$.

So, purely in terms of analysis, our problem is: given a continuous function $g:[-1,1]\to\mathbb{R}$ which is not identically $0$, prove there exists a continuous function $f:[-1,1]\to\mathbb{R}$ such that $f(0)=0$ and $\int_{-1}^1f(x)g(x)dx\neq 0$.

As a hint for how to solve this analysis problem, since $g$ is not identically $0$, there is some $a\in [-1,1]$ such that $g(a)\neq 0$. You will want to use this $a$ in your definition of the required function $f$.

• When you put it that way, it makes so much more sense. It's been a while since I've done analysis, but I think I can do it from what you and the other answers have said. Thank you so much! Apr 15 '18 at 6:20

Suppose that $g\in U^{\perp}$. Let $f$ be a function such that $f$ is almost equal to $g$ and such that $f\in U$. (You can achieve this by modifying $g$ in a neighbourhood of $0$). Then $$0=\int_{-1}^1f(x)g(x)\mathrm{d}x\sim \int_{-1}^1g(x)^2\mathrm{d}x.$$ Since we can approach $g$ arbitrarily close by $f$, we get that $g=0$.

Suppose $f\in U^{\perp}$. $\forall x_0\in[-1,1]$, if $f(x_0) \neq0$, say $f(x_0)>0$. By continuity, there is a neighborhood $V$ of $x_0$ such that $f(x) > f(x_0)/2$ in $V$. And we can always choose a non-negative function $g$ which lies in $U$ and is supported in $V$. Now $\langle f,g\rangle > 0$, which is a contradiction.