Find Continuous Functions Subject to an Integral Condition

[evinda]

Hello! (Wave)

I want to find all the continuous functions $f: [0,1] \to \mathbb{R}$ for which it holds that:$$\int_0^1 f(t) \phi''(t) dt=0, \forall \phi \in C_0^{\infty}(0,1)$$

If we knew that $f$ was twice differentiable, we could say that $\int_0^1 f(t) \phi''(t) dt= \int_0^1 f''(t) \phi(t) dt+ f(1) \phi'(1)-f(0) \phi'(0)$

What can we do if we are not allowed to differentiate $f$ ? (Thinking)

 

[Ackbach]

Try the Fundamental Lemma of the Calculus of Variations.

 

[math771]

Hi there! It seems like you are looking for some help with your math problem. As an internet forum user, I am happy to assist you in any way I can. Firstly, let's start by defining what a continuous function is. A continuous function is one that has no sudden jumps or breaks in its graph and can be drawn without lifting your pencil from the paper.

Now, let's take a look at your problem. You are trying to find all the continuous functions $f: [0,1] \to \mathbb{R}$ that satisfy the condition $\int_0^1 f(t) \phi''(t) dt=0$ for all $\phi \in C_0^{\infty}(0,1)$. This condition essentially means that the integral of $f(t)$ multiplied by the second derivative of $\phi(t)$ over the interval $[0,1]$ is equal to zero for any smooth function $\phi(t)$ that vanishes at the endpoints of the interval.

Now, if we assume that $f$ is twice differentiable, we can use integration by parts to rewrite the integral as $\int_0^1 f''(t) \phi(t) dt+ f(1) \phi'(1)-f(0) \phi'(0)$. However, since we are not allowed to differentiate $f$, we cannot use this method.

One approach we could take is to use the definition of a derivative to approximate the second derivative of $\phi(t)$ and then use the given condition to find a relationship between $f(t)$ and $\phi(t)$. Another option could be to use Taylor's theorem to approximate $f(t)$ and then use the given condition to find a relationship between the coefficients of the Taylor series.

I hope this helps you in your problem-solving process. Let me know if you have any further questions or if you need any clarification. Good luck!