To prove that on $C([0,1])$, the integral $\int_{0}^{1} f(x)g(x)dx $ defines a scalar product.

Question

Now, for the given operation to be a scalar product, I know I need to check four conditions. Here's what I have done so far:

1. $\langle f,g\rangle $ = $\int_{0}^{1} f(x)g(x)dx$ = $\int_{0}^{1} g(x)f(x)dx $ = $\langle g,f\rangle $
2. $\langle f+g,h\rangle $ = $\int_{0}^{1} (f+g)(x)h(x)dx$ = $\int_{0}^{1} f(x)h(x)dx$ + $\int_{0}^{1} g(x)h(x)dx$ = $\langle f,h\rangle $ + $\langle g,h\rangle $
3. $\langle cf,g\rangle $ = $\int_{0}^{1} (cf)(x)g(x)dx$ = c$\int_{0}^{1} f(x)g(x)dx$ = c$\langle f,g\rangle $
4. $\langle f,f\rangle $ = $\int_{0}^{1} f(x)f(x)dx$ = $\int_{0}^{1} f(x)^2dx\ \geq 0$ since, $f(x)^2 \geq 0$ $\forall x\in[0,1]$. But, now how do I show that $\langle f,f\rangle = 0 $ if and only if $f$ is the constant function zero?

Any help would be greatly appreciated.

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EDIT is it necessary to explicitly demonstrate that $\langle f,f\rangle = 0 $ if and only if $f$ is the constant function zero?

EDIT here's how I approached the problem based on the replies I got here.

Assuming $f \neq 0,$ we know that $f(a)^2 = c \geq o$ $ a \in [0,1] $ Then we have by definition: for every $\epsilon > 0 \exists \delta > 0 $ Such that $|x-a| < \delta \implies |f(x)-f(a)| < \epsilon$ and so we can get $|f(x)|> 1/2 f(a)$ or $f(x)^2> 1/4 f(a)^2$

Thus, $\int_{0}^{1} f(x)^2dx $ $\geq$ $\int_{a-\delta}^{a+\delta} f(x)^2dx $ $\geq$ $\int_{a-\delta}^{a+\delta} 1/4f(a)^2dx $ $\geq$ $1/4f(a)^2$ $\int_{a-\delta}^{a+\delta}dx$ $\geq$ $1/4f(a)^2\delta$ > 0

(Since $1/4f(a)^2 >0$ and $\delta$ > 0)

Answer 1

Of course, if $f$ is the zero function $0$, we have $$\color{#bf0000}{\langle 0, 0 \rangle} = \int_0^1 0^2 \,dx \color{#bf0000}{ = 0 }.$$

On the other hand, if $f$ is not the zero function, at some point $x_0 \in [0, 1]$ we have $f(x_0) \neq 0$, and by continuity there is some $\delta > 0$ such that $$|f(x)| \geq \frac{1}{2} f(x_0)$$ on the interval $I := [x_0 - \delta, x_0 + \delta] \cap [0, 1]$ (which has length at least $\delta$); in particular, if we denote the (positive) r.h.s. by $S$, we have $f^2(x) \geq S^2$ on that interval. Now, since $f^2$ is nonnegative, we have $$\color{#bf0000}{\langle f, f \rangle} = \int_0^1 f^2 \,dx \geq \int_I S^2 \,dx = S^2 \int_I dx \geq S^2 \delta \color{#bf0000}{ > 0}$$ as desired.

Answer 2

Hint: If $g$ is integrable, with $g(x)\geq0,$ and

$$\int_0^{1}g(x)dx=0,$$

then $g(x)=0$ in almost every point. (This is a theorem of Real Analysis.)

Answer 3

Assume $f \neq 0$, so that $f(x_0) =a>0$ (since we are integrating $f^2$. By continuity, there is an interval $(x_0- \epsilon, x_0+ \epsilon); \epsilon >0$ where $f^2>a/2$. Then $\int f^2 \geq \int_ {x_0 -\epsilon}^{x_0+ \epsilon} f^2 = 2\epsilon a >0$