Prove the existence of exactly two maxima for a positive $L^1$ function

Question

I have a function $f:\mathbb R \rightarrow \mathbb R^+$ with the following properties

• it is $L^1$ and $C^2$
• it has one single extremum (maximum) at $x=0$
• it is symmetric: $f(x)=f(-x)$.
• it is strictly positive

So there is no chance for $f(x)$ to look different from that

Now consider $$g_{\delta,a}(x)=f(x+\delta)^2+f(x-\delta)^2 - 2af(x+\delta)f(x-\delta)$$ where $\delta>0$ and $a\in [0,1]$. It is also a symmetric and strictly positive $L^1$ function.

I now want to prove that $g_{\delta,a}$ has exactly three extrema: one minimum at $x=0$ and two maxima. Of course this requires conditions for $a$. And if necessary we can make further assumtpions for $f(x)$.

It is easy to prove that there is a minimum at $x=0$ if $a<1+\frac{2f'(\delta)^2}{f(\delta)f''(\delta)}$ by just checking first and second derivative of $g_{\delta,a}$.

But then there is actually no chance for $g_{\delta,a}$ to look different from that:

But I am not able to prove that there are exactly two maxima (and not more). Or can someone give a counterexample?

Answer 1

The boundary points can be higher than the critical point itself or equal to it. So the graph may figure as a U or V. I wonder if that helps

Answer 2

Consider the following example: $$ f(x)=\frac{1}{1+x^4},\quad \delta=\frac13,\quad a=\frac34. $$ It has two minima and three maxima (one of then at $x=0$.) There is a mistake in your computations. The condition for $g''_{\delta,a}(0)>0$ is $$ (1+a)f'(\delta)^2+(1-a)f(\delta)f''(\delta)>0. $$