Approximating $L^1$ function with convolutions

Question

Let $f\in L^1([0,1])$, and suppose that $\int_0^1f\varphi^{(n)} = 0$ for every $\varphi\in C_c^\infty(0,1)$, where $\varphi^{(n)}$ is the nth derivative. Show that $f$ is a polynomial of degree at most $n-1$
(Given hint: Approximate $f$ by smooth functions using convolutions)

Thoughts:
Going off the hint, we choose $\psi\in C_c^\infty$ such that $\int_0^1\psi =1$ and define $\psi_\epsilon(x) = \frac{1}{\epsilon}\psi(x/\epsilon)$ so that $(f*\psi_\epsilon)\to f$ in $L^1$

After this, I'm a bit unsure.
If we somehow show $\int_{[0,1]} (f*\psi_\epsilon)^{(n)}\varphi = 0$ (which I haven't), and letting $\epsilon \to 0$, how would that show $f$ is even differentiable, and a polynomial.

Answer 1

The trick is to move the convolution from $f$ to $\varphi$. Just to have the definition well posed we assume $f,\ \varphi$ and $\psi$ to be defined as $0$ outside $[0,1]$ (convolution is defined on the whole space). $$\int_0^1 (f*\psi_\epsilon)^{(n)}(x)\varphi(x)dx=(-1)^{n}\int_0^1 (f*\psi_\epsilon)(x)\varphi^{(n)}(x)dx=\\ (-1)^{n}\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}f(y)\psi_\varepsilon(x-y)\varphi^{(n)}(x)dydx=\\ (-1)^{n}\int_{-\infty}^{+\infty}f(y)\int_{-\infty}^{+\infty}\psi_\varepsilon(x-y)\varphi^{(n)}(x)dxdy=(-1)^{n}\int_{-\infty}^{+\infty}f(y)(\tilde{\psi}_{\varepsilon}*\varphi^{(n)})(y)dy=\\ (-1)^{n}\int_{0}^{1}f(y)(\tilde{\psi}_{\varepsilon}*\varphi^{(n)})(y)dy= (-1)^{n}\int_{0}^{1}f(y)(\tilde{\psi}_{\varepsilon}*\varphi)^{(n)}(y)dy=0, $$ where $\tilde{\psi}(t):=\psi(-t)$. Just to elaborate a little bit:

• The second equality holds because the support of $\varphi^{(n)}$ is contained in $(0,1)$.
• The third equality from the bottom holds for $\varepsilon$ small enough because the support of $\tilde{\psi}_{\varepsilon}*\varphi^{(n)}$ is contained in an $\varepsilon-$neighborhood of the support of $\varphi^{(n)}$ which is a compact contained in $(0,1)$.
• The second equality from the bottom is just one of the properties of the convolution.
• The last equality comes from the assumption because $(\tilde{\psi}_{\varepsilon}*\varphi)^{(n)}\in C_c^\infty(0,1)$ (again for $\varepsilon$ small enough)