Does this inner product on $L^1([0,1])$ have a name? Math people:
For $f, g \in L^1([0,1])$, define
$$\langle f,g \rangle = \int_0^1 \int_0^1 f(t)g(t')\exp(-|t-t'|)dt'\,dt.$$
Although we don't normally think of $L^1([0,1])$ as an inner product space, this is an inner product on $L^1([0,1])$.  The only requirement that is not trivial to verify is positive-definiteness: $\langle f, f \rangle > 0$ for $f \neq 0$. This requires a straightforward argument that leads to integration by parts.  
My question is, has this been done before?  Does this have a name?  You can use an interval other than $[0,1]$ and use a sum of several positively weighted exponential functions if you want, but I wanted to give the simplest possible example.
 A: We can ask which kernels $K(x,y)$ (with symmetry property $K(x,y)=K(y,x)$ or hermitian symmetry, if complex-valued) are "positive-definite" in the sense that $\int\int f(x)\overline{f(y)}K(x,y)\,dx\,dy\ge 0$.
Thinking of the finite-dimensional (=matrix) analogue, certainly positivity of entries is neither necessary nor sufficient to guarantee positive-definiteness. Rather, the diagonal entries must sufficiently "dominate" the other entries (and be positive themselves). 
The simplest case would be that $K(x,y)$ is $\delta_{x-y}$, that is, Dirac delta along the diagonal (thinking in terms of Schwartz kernel theorem). This is the kernel for the identity map, indeed, and $\int f(x)\overline{f(x)}\;dx\ge 0$ as desired.
The example at hand, and a variety of others, are "Green's functions" for "definite" differential operators: here, $((d/dx)^2-1)e^{-|x-y|}=-2\delta_{x-y}$. Integration by parts shows that $L=((d/dx)^2-1)$ is a negative (unbounded) symmetric operator in the sense that $\int Lf(x)\overline{f(x)}\,dx\le 0$. Thus, its inverse is also negative-definite, and
$$
\int\int f(x)\,\overline{f(y)}\,L_x^{-1}\delta_{x-y}\;dx\,dy
\;=\;\int\int L^{-1}f(x)\,\overline{f(y)}\,\delta_{x-y}\;dx\,dy
\;=\; -2\int L^{-1}f(x)\,\overline{f(x)}\,dx \;\ge\; 0
$$
The operator $u\rightarrow u''-u$ can be replaced by $Lu=u''-qu$ for $q>0$, for example, or any Sturm-Liouville type operator $(pu')'-qu$ with $p>0$ and $q>0$. Then the negative of a Green's function for $L-c$ for any $c>0$ will have the same positivity property, by the same argument. Higher-order differential operators achieve similar effects.
Further, thinking about domination by the diagonal, given a positive symmetric kernel constructed as just above, obviously perturbing by demonstrably-smaller kernels does not harm positivity. From the other side, in many higher-dimensional examples, we don't have to have a Green's function, but just a suitable parametrix (for positive/negative-definite elliptic operator).
(Sorry for earlier dumb remarks!)
As @ABlumenthal notes, however, this will definitely not give the usual topology on $L^1[0,1]$, since it will give a Hilbert-space topology.
