Let f be continuous on [a,b] and suppose that, for every integrable function g defined on [a,b],∫_{a}^{b}fg=0. Prove that f(x)=0 for all x∈[a,b].
Why isn't g(x)= 0 why does it have to be f(x)? Why not both?
f is continuous and g is integrable and the product of the integral of fg over the interval [a,b] is 0. So f(x) must be 0 for all x in [a,b], is all the information I have been given. I will start with the definition of continuous for f and integrable for g and then hope I see something useful after that.
I came up with the following solution, I hope it is not so wrong.
Suppose that $\displaystyle f$ is not $\displaystyle 0$ for all $\displaystyle x\in[a,b]$, then there exists $\displaystyle x_0\in[a,b]$ such that $\displaystyle f(x_0)>0$ (If the is no such $\displaystyle x_0$, there must be $\displaystyle x_1\in[a,b]$ such that $\displaystyle f(x_1)<0$.In that case we will look at $\displaystyle -f$).
f is continuous function therefor exists $\displaystyle \delta>0$ for which $\displaystyle f(x)>\epsilon>0$ for all $\displaystyle x\in (x_0-\delta,x_0+\delta)$.
Now we define $\displaystyle h(x)$-'staircase'-characteristic function of open interval $\displaystyle (x_0-\delta,x_0+\delta)$.
$\displaystyle h(x)=\left\{\begin{matrix}1 & x\in (x_0-\delta,x_0+\delta) \\ 0 & \text{otherwise} \end{matrix}\righ$
Hence, we will get:
$\displaystyle 0=\int_{a}^{b}f(x)h(x)dx=\int_{a}^{x_0-\delta}f(x)h(x)dx+\int_{x_0-\delta}^{x_0+\delta}f(x)h(x)dx+\int_{x_0+\delta}^{ b}f(x)h(x)dx=0+\int_{x_0-\delta}^{x_0+\delta}f(x)h(x)dx+0>2\epsilon\delta $.
So, we got a contradiction to the fact that if $\displaystyle f$ is continuous on $\displaystyle [a,b]$ and h is staircase function $\displaystyle \int_{a}^{b}f(x)h(x)dx=0 \Rightarrow f\equiv 0$.
But we also know that if $\displaystyle g$ is integrable function on $\displaystyle [a,b]$, so for every $\displaystyle \epsilon_0>0$ exists staircase function $\displaystyle h$ so that
$\displaystyle \int_{a}^{b}|g(x)-h(x)|<\epsilon_0$
Combining these two results we will conclude the needed.
Your solution is correct.
Note that $\displaystyle f \text{ is continuous on }[a,b]$ so as such that means that both $\displaystyle f~\&~f^2$ is are integrable on $\displaystyle [a,b]$.
By the given $\displaystyle \int_a^b {f \cdot f} = \int_a^b {f^2 } = 0$.
But by your reasoning that must imply that $\displaystyle f \equiv 0$.