Does this operator have a continuous, localized eigenfunction with negative eigenvalue? I am looking at a class of operators
$$
L[f](x)=af_{xxxx}-bf_{xx}+\frac{d}{dx}(\delta(x)f_x)
$$ , a<0,b<0,
on the real line, where $\delta$ is Dirac-delta.
I am interested in ruling out the possibility that this operator has a continuous localized eigenfunction with negative Eigenvalue.
Localized means it goes to 0 as $|x|\rightarrow \infty$.
Away from origin, a candidate is
$$
v(x)=c\exp(-k|x|)
$$ for some appropriate $k$. However, when looking at its behavior near origin, if I compute
$$
\int_{-\epsilon}^{\epsilon}\frac{d}{dx}(\delta(x)v_x)dx=\frac{1}{2\epsilon}(v_x(\epsilon)-v_x(-\epsilon))=\frac{1}{2\epsilon}(2c),
$$ the term blows up as $\epsilon\rightarrow 0$ Hence we conclude no continuous localized eigenfunctions exist. Here I have approximated Dirac-delta by a constant function of width $w=2\epsilon$ and height $1/w$.
On the other hand, if I consider a discontinuous candidate
$$
v(x)=\operatorname{sign}(x)\cdot c\exp(-k|x|),
$$
the limit of the same integral is now = $ \frac{1}{2\epsilon}(v_x(\epsilon)-v_x(-\epsilon))=\frac{1}{2\epsilon}(c-c)= 0$. Hence, this candidate is a localized eigenfunction.
Is the above logic accurate or am I missing something ? What do i need to make the above rigorous ?
**Some more details upon answer from Johannes Hahn and Michael Renardy
In general, $k\in \{ \pm \sqrt{\frac{b\pm\sqrt{b^2+4a\lambda}}{2a}} \}$, and solution  $f$ has the form
$$f(x) = \begin{cases} \sum_{i=1}^4 c_i e^{k_i x} & x<0 \\ \sum_{i=1}^4 d_i e^{k_i x} & x>0\end{cases}$$, where we can solve independently for $x<0$ and $x>0$.
However, we need $\lambda<0$. By looking at the expression for $k$ above, we see that since $a<0,b<0,\lambda<0$, $\sqrt{b^2+4a\lambda}>|b|$, which implies that
two of the $k$s will be imaginary (for both positive and negative $x$).
Let $r=k^2$. Then $r_1=\frac{b-\sqrt{b^2+4a\lambda}}{2a}>0$ and $r_2=\frac{b+\sqrt{b^2-4a\lambda}}{2a}<0$.
Hence, $k_{1,2}=\pm \sqrt{r_1}$ and $k_{3,4}=i\sqrt{-r_2}$.
However localization requires that only real $k$ terms survive since imaginary $k$ terms will give rise to $cos$ and $sin$ that don't decay at infinity. This implies $c_3=c_4=d_3=d_4=0$. Hence the candidate solution is reduced to:
$$f(x) = \begin{cases} \sum_{i=1}^2 c_i e^{k_i x} & x<0 \\ \sum_{i=1}^2 d_i e^{k_i x} & x>0\end{cases}$$
Furthermore, only positive $k$ is allowed for $x<0$ and negative $k$ for $x>0$ otherwise solution blows up as $|x|\rightarrow \infty$. This further implies $c_2=d_1=0$.  This leaves us with only two options as mentioned earlier:
Continuous candidate: $\cdot c\exp(-k|x|)$,
and
Discontinuous candidate: $\operatorname{sign}(x)\cdot c\exp(-k|x|)$, where $c=c_1=d_2.$
 A: If $f$ is smooth, then $\delta \cdot f = f(0)\delta$ and consequently $(\delta \cdot f')' = f'(0) \delta'$. If $f$ is at least $C^1$ we can take this as the definition of the term. If $f$ is not $C^1$ around zero, then what does $\delta\cdot f'$ even mean?
Anyway, if it is at least $C^1$, then we can rearrange the eigenfunction equation
$$\lambda f = af^{(4)}-bf'' + f'(0)\delta'$$
to $f'(0)\delta' = -af^{(4)} + bf''+\lambda f$. Because, you're interested in the behaviour of $f$ at infinity, we can simply solve the ODE $0=-af^{(4)} +bf'' + \lambda f$ away from zero. Such an $f$ is an arbitrary linear combination of the functions $e^{k x}$ with $k$ one of the roots of $aX^4 - bX^2 - \lambda =0$, i.e. $k\in \{ \pm \sqrt{\frac{b\pm\sqrt{b^2+4ab\lambda}}{-4a}} \}$ (at least generically. For special values of $\lambda$, the general solution will look different)
We get to choose the coefficients of this linear combination independently on the right and left half of the real line, so that $f$ has the form
$$f(x) = \begin{cases} \sum_{i=1}^4 c_i e^{k_i x} & x>0 \\ \sum_{i=1}^4 d_i e^{k_i x} & x<0\end{cases}$$
for some $c_i, d_i$ of our choosing. In order for $f$ to be continuous, our choices have to satisfy $\sum_{i=1}^4 (c_i-d_i)=0$. In order for $f$ to be $C^1$ around zero, our choices have to satisfy $\sum_{i=1}^4 k_i (c_i-d_i)=0$. In fact, since we have eight degrees of freedom, we can make $f$ at least $C^4$ and also satisfy $f'(0)=0$ which makes the $\delta'$ term disappear and we have found an eigenfunction for the original operator.
A: Yes, such eigenfunctions exist.
If $a>0$, $b<0$ and $\lambda<0$, the roots $k$  of $ak^4-bk^2-\lambda=0$ are complex. Let $k_1$, $k_2$ be the roots with positive real part. Assume they are different; we can modify the calculation for the case when they are not. We are looking for an odd eigenfunction so that
$$f=c_1e^{-k_1x}+c_2e^{-k_2x},\, x>0,$$
$$f=-c_1e^{k_1x}-c_2e^{k_2x},\,x<0.$$
At the origin, we need $f=0$ and $a(f_{xx}(0+)-f_{xx}(0-))+f_x(0)=0$. This leads to
$$c_1+c_2=0,\,c_1(2ak_1^2-k_1)+c_2(2ak_2^2-k_2)=0.$$
For nontrivial solutions to exist, we require $2ak_1^2-k_1=2ak_2^2-k_2$. For $k_1\neq k_2$, this leads to $2a(k_1+k_2)=1$.
We also have $ak_1^4-bk_1^2=ak_2^4-bk_2^2$, which leads to $a(k_1^2+k_2^2)=b$. We now have two equations for $k_1$ and $k_2$, which yield complex solutions with positive real part as desired. Multiplying the last equation by $k_1^2$, we find $ak_1^4-bk_1^2=-ak_1^2k_2^2=\lambda$. Since $a$ is positive, and $k_1$ and $k_2$ are conjugates, this is indeed negative.
In the above analysis, I assumed $a>0$, $b<0$ and $\lambda<0$ as originally stated. Not all comments made here are consistent with this, so I cannot be sure this is what was intended.
