Find $a>0$ and $f \in C[0,1]$ such that $f \neq 0$, $f(1) = 0$ and $\int_0^1 \cosh(ax) f(x) dx = 0$

Question

Do there exist $a>0$ and $f \in C[0,1]$ such that

• $f$ is not $0$ everywhere,
• $f(1) = 0$
• $\int_0^1 \cosh(ax) f(x) dx = 0$?

Here $\cosh$ is hyperbolic cosine, that is $$ \cosh(x) = \frac{e^x+e^{-x}}{2}. $$

Of course the answer is no if we consider only nonnegative $f$.

Answer 1

For example, you might try $f$ of the form $f(x) = b (1-x) + c (1-x)^2$. I get

$$ \int_0^1 \cosh(ax) f(x)\; dx = {\frac {\cosh \left( a \right) ab+2\,c\sinh \left( a \right) -ba-2\,ca }{{a}^{3}}} $$

so for any $a \ne 0$ and $b \ne 0$ you can take

$$ c = -{\frac {a \left( \cosh \left( a \right) -1 \right) }{2 (\sinh \left( a \right) -a)}} b$$

More generally, for any linearly independent $f_1, f_2 \in C[0,1]$ with $f_i(1)=0$, let $g_1(a) = \int_0^1 \cosh(ax) f_1(x)\; dx$ and $g_1(a) = \int_0^1 \cosh(ax) f_2(x)\; dx$. Then $f(x) = b f_1(x) + c f_2(x)$ will work if $b g_1(a) + c g_2(a) = 0$, and you can take $b$ and $c$ not both $0$ to make this true.

Answer 2

Consider $a=1$ and$$f(x)=\operatorname{sech} x\sin 2\pi x\quad(0\leqslant x\leqslant1).$$The first factor makes the integration easy, and the second is chosen to make the integral zero while satisfying the end condition $f(1)=0$.