Maximum value of $\int_{0}^{1}e^x\log f(x)dx$ when $\int_{0}^{1}f(x)dx=1$ Suppose that $f(x)\ (0\le x\le 1)$ is continuous and strictly positive and satisfies $$\int_{0}^{1}f(x)dx=1.$$
Then, can we find the maximum and the minimum value of the following? If yes, then how?
$$\int_{0}^{1}e^x\log f(x)dx$$
I even don't know if this is solvable. Is there any theorem? 
 A: There is no lower bound. Pick $f(x) = (n+1)x^n$. Then $f(x)$ satisfies the hypotheses except at the origin (so one can simplify modify the function in a small neighborhood of the origin later), and
$$ \int_0^1 e^x \log(f(x)) \, dx = (e - 1) \log(n+1) + n \int_0^1  e^x \log{x} \, dx$$
We can observe that
$$ \int_0^1 e^x \log{x} \, dx < 0$$
So the first term is growing logarithmically, while the second term is decreasing linearly, so for arbitrarily large $n$, the integral will be arbitrarily negative.
To find the maximum, first pick $f(x) = e^x / (e - 1)$. Then one finds that
$$ \int_0^1 e^x \log(f(x)) \, dx = 1 - (e - 1) \log(e - 1) \approx 0.07$$
I claim that this is the maximizer. To do this, define the functional
$$ I(f) = \int_0^1 e^x \log(f(x)) \, dx $$
If $f_0(x) = e^x/(e - 1)$, then any admissable $f(x)$ can be written as $f(x) = f_0(x) + \epsilon g(x)$, where $g(x)$ is some function such that $\int_0^1 g(x) \, dx = 0$. Then
$$ \frac{d}{d\epsilon} I(f_0 + \epsilon g) = \int_0^1 e^x \frac{g(x)}{f_0(x) + \epsilon g(x)} \, dx $$
Therefore, by setting $\epsilon = 0$, the directional derivative of $I$ at $f_0$ in the direction of $g$ is given by
\begin{align*}
\frac{d}{d\epsilon} \left. I(f_0 + \epsilon g) \right|_{\epsilon = 0} & = \int_0^1 e^x \frac{g(x)}{f_0(x)} \, dx \\
& = \int_0^1 (e -1) g(x) \, dx \\
& = 0.
\end{align*}
The derivative vanishes independent of $g$, and hence $f_0$ is a local extrema of $I$. Calculating the second derivative,
\begin{align*}
\frac{d^2}{d\epsilon^2} I(f_0 + \epsilon g) & = - \int_0^1 e^x \frac{g(x)^2}{(f_0(x) + \epsilon g(x))^2} \, dx \\
& < 0,
\end{align*}
and hence $f_0(x)$ is at least a local maximum. However, the above calculation of the second derivative is valid even when $f_0(x)$ is replaced by any other admissable function $f(x)$, and hence this shows that $I$ is in fact strictly concave. Therefore $f_0(x)$ is the global maximum.
A: My try: for $n \in \mathbb{N}$ consider $(n+1)x^n$. We have $\int_0^1 (n+1)x^n \ dx =1$.
Then
$$ \int_0^1 e^x \log ((n+1)x^n) \ dx = \int_0^1 e^x (\log (n+1) + n \log x) \ dx = $$
$$= (e-1) \log(n+1) + n \int_0^1 e^x \log x \ dx = (e-1) \log(n+1) + Cn$$
Where $C=-1.3\dots$ approximatively.
So, taking $n \longrightarrow + \infty$ you get $$\int_0^1 e^x \log ((n+1)x^n) \ dx \longrightarrow - \infty$$
Now one can approximate these functions (which are not strictly $>0$) with $$\frac{(n+1)x^n + \varepsilon}{1+ \varepsilon} >0$$
for $\varepsilon >0$,
and get that your functional at least has no minimum.
A: Use the inequality $$1-\frac {1}{u}>\log (u)>u-1$$ valid for all $u\in (0,1)$, (Moreover,  when $ u > 1$ the the inequality is reversed.                                  
Case (i):   If $ 0<f (x)<1$,                                                 Then, $$1-\frac {1}{f (x)}>\log (f (x))> f (x)-1$$ Multiplying, each side of the inequality by the positive function $ e^x $ and then integrating each side with respect to $ x $ from $0$ to $1$ to get $$ \int_0^1 {e^x (1-\frac {1}{f (x)})dx} >I:=\int_0^1 {e^x \log (f (x))dx}>\int_0^1 {e^x (f (x)-1) dx} $$ Now, for the left  hand integral we have $$\int_0^1 {e^x (1-\frac {1}{f (x)})dx} <e\cdot \int_0^1 {1-\frac {1}{f (x)}dx}<e+\frac {e}{M} $$ where we used the fact that since $ f $ is continuous on the $[0,1] $ then there exists $ M > 0$ s. t. $-M <f (x)<M $. In a similar manner we may observe that for the right hand integral we have $$\int _0^1 {e^x (f (x)-1)dx }> 0$$ Combining all inequalities to get $$e+\frac {e}{M}> I> 0$$ finally, letting $ M →\infty $ we get $0 <I <e$.                                                                                        
Case (ii): if $f (x)> 1$. Then, using the same techniques above with reversed inequality we get $1 <I <e-1$.
A: Take $f(x) = \epsilon$ if $x\in [1/2,1)$ and $f(x) =4(1-\epsilon) (1-2x) + \epsilon$ if $x\in (0,1/2)$ then $\int_0^1 f(x) dx =1$ and
$$\int_0^1 f(x) \ln f(x) dx =\int_0^{1/2} e^x \ln (4(1-\epsilon)(1-2x) + \epsilon) dx +  (e -e^{1/2})ln \epsilon \to \int_0^{1/2} e^x \ln (4(1-2x)) dx + -\infty =-\infty$$
as $\epsilon \to 0$. Thus the infimum is $-\infty$.
Using the inequality $ab \leq e^a + b\ln b -b$ for any $a\in \mathbb R$ and $b >0$, we get
$$\frac{e^x}{e-1} \ln f(x) \leq e^{\ln f(x)} + \frac{e^x}{e-1}\ln \frac{e^x}{e-1} -\frac{e^x}{e-1} =f(x) + \frac{e^x}{e-1}(x-1) -\frac{e^x}{e-1} \ln (e-1)$$
which implies
$$\int_0^1 e^x \ln f(x) dx \leq 1 -(e-1)\ln (e-1).$$
Thus the maximum value is $1 -(e-1)\ln (e-1)$ and is attained by function $e^x/(e-1)$.
