Functional Equations and Integration

No such function with the property that $f(1+x)+f(1-x)=e^x$ for all real $x$ can exist.

Substitute $x=1$ and $x=-1$ : $\begin{aligned} f(2)+f(0)&=e\\ f(0)+f(2)&=e^{-1} \end{aligned}$ which implies $e=e^{-1}$ , which is impossible.

Therefore, the function $f$ does not exist.

In fact, different ways to approach this problem will result different results, which should be a flag that something is amiss. For example, $\begin{aligned} \int_0^1f(1+x)\ dx+\int_0^1f(1-x)\ dx&=\int_0^1e^x\ dx\\ \int_0^2f(x)\ dx&=e-1 \end{aligned}$

but if one instead tries substituting $u=1-x$ : $\begin{aligned} f(2-u)+f(u)&=e^{1-u}\\ \int_0^2f(2-u)\ du+\int_0^2f(u)\ du&=\int_0^2e^{1-u}\ du\\ 2\int_0^2f(u)\ du&=1-\frac{1}{e}\\ \int_0^2f(u)\ du&=\frac{e-1}{2e}\\ \end{aligned}$

Let $u = 1+x, v = 1-x$ so that $\frac{u-v}{2} = \frac{(1+x)-(1-x)}{2} = x.$ Also, when $x = 0$ we obtain the condition $f(1) + f(1) = 2f(1) = 1 \Rightarrow f(1) = \frac{1}{2}.$ Substituting these into the original functional above yields:

$f(u) + f(v) = e^{\frac{u-v}{2}}$ (i)

Differentiating (i) with respect to u and v respectively gives:

$f'(u) = \frac{1}{2} e^{\frac{u-v}{2}};$ and $f'(v) = -\frac{1}{2} e^{\frac{u-v}{2}};$ , which gives $f'(u) = -f'(v) = A$ for $A \in \mathbb{R}.$ Solving this first-order ODE then yields:

$f(u) = Au + B$

and utilizing the initial condition $f(1) = \frac{1}{2}$ finally produces $f(u) = A(u-1) + \frac{1}{2}$ (ii). Replacing (ii) back into original functional equation now gives:

$f(1+x) + f(1-x) = A(1+x-1) + \frac{1}{2} + A(1-x-1) + \frac{1}{2} = Ax - Ax + 1 \Rightarrow \boxed{1 = e^{x}}.$ But the functional equation must hold true for ALL $x \in \mathbb{R} \Rightarrow$ no such function $f(x)$ exists!

Differentiate the given expression with respect to x and evaluate the resulting expression for x=0. The result of this will give 0=1, which is obviously impossible.