# Question about differentiating under the integral sign

Let $0<m(E)<\infty$ for a Lebesgue measurable subset $E \subset \mathbb{R}$. Consider two positive functions $f$ and $g$ in $L^p(E)$, $p>1$, and define the function $F:[0,\infty) \to [0, \infty)$ as $$F(t)= \int_E [f(x)+tg(x)]^pd x.$$ Show $F$ is differentiable on $(0, \infty)$ and compute its derivative $F'(t)$. Make sure to show $F'(t)$ is finite.

This came from an old qualifying exam at my university. Here is my attempt (letting $h(x,t)$ stand for the function in the integral):

Since $p-1>0$ and since $f$ and $g$ are positive, the partial $$\frac{\partial h}{\partial t}=pg(x)[f(x)+tg(x)]^{p-1}$$

exists for each $(x,t), x\in E, t\in [0,\infty)$. Also note for each $(x,t)$ again since $g$ and $f$ are positive, \begin{align*} \bigg|\frac{\partial h}{\partial t}\bigg|&=\big|pg(x)[f(x)+tg(x)]^{p-1}\big| \\ & \leq p 2^{p-1}\big(f(x)^{p-1}g(x)+t^{p-1}g(x)^p\big) \end{align*}

The finiteness of $m(E)$ ensures that the latter function is integrable. Hence, $$\frac{\partial F}{\partial t}=\frac {\partial}{\partial t} \int_E h(x,t)\:\mathrm{d}x=\int_E \frac{\partial h}{\partial t}\:\mathrm{d}x=p\int_Eg(x)[f(x)+tg(x)]^{p-1}\:\mathrm{d}x < \infty$$ for all $(x,t)$ as previously described.

I feel uneasy with this solution, particularly with the inequalities, and whether I correctly justified differentiation under the integral sign.

You should name the nontrivial theorems that you are using if you want us to judge whether you are justifying everything correctly. This advice extends to the quals if you are taking them in the future, and beyond.

For the first inequality, you can indeed start with $f + tg \leq 2 \max(f, tg)$. But don't take the sum just yet – instead, take the power first, whence $$(f + tg)^{p-1} \leq 2^{p-1}[\max(f, tg)]^{p-1} = 2^{p-1} \max(f^{p-1}, t^{p-1}g^{p-1}) \leq 2^{p-1}(f^{p-1}+t^{p-1}g^{p-1}).$$

Next, for the second inequality and when you write "the latter function is integrable", are you possibly trying to argue that the product of two $\mathscr L^1$ functions is $\mathscr L^1$? If so, don't: It's not true in general. Instead, applying Hölder on the $f^{p-1}g$ term gives $$\lVert f^{p-1}g \rVert_1 \leq \lVert f^{p-1} \rVert_{p/(p-1)} \lVert g \rVert_p = \lVert f \rVert_p^{p-1} \lVert g \rVert_p < \infty$$ where we used $$\lVert f^{p-1} \rVert_{p/(p-1)} = \left(\int (f^{p-1})^{p/(p-1)}\right)^{(p-1)/p} = \left(\int f^p\right)^{(p-1)/p} = \lVert f \rVert_p^{p-1}.$$

And for the derivative under the integral, you may be able to find a ready-made theorem in the textbook of your (school's) choice. Nonetheless, it would be safer to know how to justify the exchange using the mean value and the dominated convergence theorems. We argue as follows: On any bounded open interval $I \subset (0, \infty)$ containing $t$, the mean value theorem tells us that for all $s \in I \setminus \{t\}$, we have \begin{align*} \left\lvert \frac{h(x,s) - h(x,t)}{s - t} \right\rvert &= \lvert \partial_t h(x,c_s) \rvert \\ &\leq \sup_{c \in I} \lvert \partial_t h(x,c) \rvert \\ &\leq p2^{p-1} \left( f(x)^{p-1} g(x) + (\sup I)^{p-1} g(x)^p \right) \in \mathscr L^1 \end{align*} where $c_s \in I$ depends on $s$. Thus the dominated convergence theorem allows us to move the limit in the definition of $F'$ inside the integral, giving $F' = \int \partial_t h \, \mathrm{d}x$.

We can in fact skip the first inequality. Since $f+tg \in \mathscr L^p$, applying Hölder directly gives $$\lVert (f+tg)^{p-1}g \rVert_1 \leq \lVert (f+tg)^{p-1} \rVert_{p/(p-1)} \lVert g \rVert_p = \lVert f+tg \rVert_p^{p-1} \lVert g \rVert_p < \infty.$$ Hence $\partial_t h(x,t)$ is integrable over $x \in E$ for all $t$. More generally, we have that weighted geometric means of nonnegative real $\mathscr L^p$ functions are also $\mathscr L^p$.

I feel the approach you have taken is correct but I don't understand your inequalities. Instead, I propose the following inequalities.

\begin{align*} \bigg|\frac{\partial h}{\partial t}\bigg|&=\big|pg(x)[f(x)+tg(x)]^{p-1}\big| \\ & \leq p \left( \frac{\lvert g(x) \rvert^p}{p} + \frac{\lvert f(x)+tg(x)\rvert^p}{q}\right) \end{align*}

Here, I used Young's inequality : $\vert ab \vert \leq \frac{a^p}{p} + \frac{b^q}{q}$ where $q = \frac{p}{p-1}$ Substitute $a= g(x)$ and $b = [f(x)+tg(x)]^{p-1}$

The integral of right side of the inequality above (with respect to $x$) is finite as both $f$ and $g$ are $p$ integrable. Hence, from Lebesgue Dominated Convergence Theorem we can swap integral and differentiation and we have the derivative as you calculated.

• Good point! I meant, by my own inequalities, that $$[f(x)+tg(x)] \leq 2\max\big(f(x),tg(x)\big)\leq 2\big(f(x)+tg(x)\big)$$ as all quantities involved are positive (assuming $t=0$ trivially works as well). Since $p-1>0$ and both sides of the above inequality are positive, we then have $$[f(x)+tg(x)]^{p-1}\leq 2^{p-1}\big(f(x)+tg(x)\big)^{p-1}$$ since $\xi^{p-1}$ is strictly increasing on $\xi>0$. This overdoes things if $p-1 \geq 1$ but it still works, and furthermore takes care of the case if $0<p-1<1$. The finiteness of $E$ guarantees that the $L^p$ functions are in $L^{p-1}$. Correct? – Darrin Dec 27 '14 at 22:44
• $(f(x) + t g(x)) \leq 2(f(x)+tg(x))$ as $f$ and $g$ are positive. (You don't need an intermediate inequality containing max term :-)) Yes, finiteness of $E$ guarantees $L^p$ are in $L^{p-1}$ for p>1 – chandu1729 Dec 28 '14 at 6:31