Prove that for all $g$ continuous functions, with $g(0) = g(1) = 1$ and $\int f(x)g(x)dx = 0$ then $f(x) = 0$

calculusintegrationreal-analysis

I am trying to prove that if $f:[0,1] \to \mathbb{R}$ is a continuous function for which the following is true: for all continuous functions $g:[0,1] \to \mathbb{R}$ with $g(0) = g(1) = 1$, with $\int_0^1f(t)g(t)dt = 0$ then $f(x) = 0, \forall x \in [0,1]$. Any ideas on how I can approach this? I have tried playing around with g(x), and also tried approaching it more theoretically, starting by supposing that $f(x) \neq 0$ but both without much success. The problem is from a university exam on Calculus 1, where integrals are studied on a basic level (Riemann integrals, Integration techniques and basic theorems for Intergrals) . Thanks for any help!

Best Answer

Sketch.

  1. From the given hypothesis, deduce that the following is true as well: Given any continuous function $h$ with $h(0) = h(1) = 0$, we have $$\int_0^1 f(t)h(t)\ {\mathrm d}t = 0.$$ (Hint: Write $h = (h + 1) - 1$.)

  2. Assume $f$ is not identically $0$. Wlog, $f > 0$ at some point.
    Use continuity to conclude that there exists $\delta > 0$ and some interval $(a, b) \subset (0, 1)$ such that $f(x) > \delta$ for all $x \in (a, b)$.

  3. Define a continuous function $h$ on $[0, 1]$ which has the property that $h = 0$ outside $(a, b)$ and $h = 1$ on some subinterval $[c, d] \subset (a, b)$. (Try to find an $h$ with a "trapezoidal graph".)

  4. Show that $\int_0^1 fh > 0$. This is contradiction since $h$ is continuous and satisfies $h(0) = h(1) = 0$.

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