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Define $L: C^2[0,m]\to C[0,m]$, where $\displaystyle L = -\frac{d^2u}{dx^2}$.

Suppose $f \in C[0,m]$ satisfies $\displaystyle \int^m_0f(x)dx=0$.

Show that $f$ is in the range of $L$.

Proof:

Suppose $f\in C[0,m]$. So $\displaystyle-\frac{d^2u}{dx^2}=f(x)$. Integrating from $0$ to $m$, $-\displaystyle\int^m_0\frac{d^2u}{dx^2}dx=\int^m_0f(x)dx.$

Evaluating the LHS, $-\displaystyle\int^m_0\frac{d^2u}{dx^2}dx=-\left[\frac{du}{dx}\right]^m_0=-\frac{du}{dx}(m)+\frac{du}{dx}(0)=0.$

Since the $f$ in the RHS is such that $\displaystyle\int^m_0f(x)dx=0$, $f$ is in the range of $L$. QED

From what I understand, by the way $L$ is defined, any $u\in C^2[0,m]$ I take, when I integrate from $0$ to $m$ must actually evaluate to $0$. Since $f$ is defined in such a way, it shows that it is in the range of $L$. Is my reasoning valid?

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    The $\int f = 0$ constraint seems unnecessary. – 2017-01-08

1 Answers 1

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I don't see how you conclude that $f $ is in the range of $L $. You could simply take $$ u (x)=-\int_0^x\int_0^t f (s)\,ds\,dx $$ and then $Lu=f $.