Let $f \in L^2(X)$ such that $f$ is generated by some arbitrary constant; that is, $f = g(a)$ with $g: \mathbb{R} \to L^2(X)$. Then what can be said about the derivative with respect to some arbitrary variable of the inner product of $f$ with itself: $$D =\frac{\partial}{\partial a}\langle f, f \rangle = \;?$$

I had the idea that $D = 2 \langle f , \frac{\partial f}{\partial a} \rangle$, where $I$ is the identity function, but I'm not sure. The evidence for this is as follows \begin{equation} \begin{aligned} D &=\frac{\partial}{\partial a} \int_X \left(f(x)\right)^2\ dx\\ &= 2\int_X f(x) \frac{\partial f}{\partial a} \ dx. \end{aligned} \end{equation}

My ultimate goal is to compute this for general vector spaces not just $L^2$. Thanks.

  • 3
    $\begingroup$ Related, I think: math.stackexchange.com/questions/96265/…. I find your question confusing, and suspect it is because of notation. You mean you have a function from $\mathbb R$ to $L^2$, $a\mapsto g(a) = f$? In that case see the link. $\endgroup$ – Jonas Meyer Jul 16 '15 at 19:08
  • $\begingroup$ @JonasMeyer I've used your answer to answer my own question specifically aimed at the $L^2$ inner product norm. Thank you again. $\endgroup$ – MadcowD Jul 16 '15 at 19:24

Using information from Differentiating an Inner Product, one can approach this problem from a more general perspective.

If $V$, $W$, and $Z$ are normed spaces, and if $T:V\times W\to Z$ is a continuous (real) bilinear operator], meaning that there exists $C\geq 0$ such that $\|T(v,w)\|\leq C\|v\|\|w\|$ for all $v\in V$ and $w\in W$, then the derivative of $T$ at $(v_0,w_0)$ is $DT|_{(v_0,w_0)}(v,w)=T(v,w_0)+T(v_0,w)$. (I am assuming that $V\times W$ is given a norm equivalent with $\|(v,w)\|=\sqrt{\|v\|^2+\|w\|^2}$.) This follows from the straightforward computation

$$\frac{\|T(v_0+v,w_0+w)-T(v_0,w_0)-(T(v,w_0)+T(v_0,w))\|}{\|(v,w)\|}=\frac{\|T(v,w)\|}{\|(v,w)\|}\leq C\frac{\|v\|\|w\|}{\|(v,w)\|}\to 0$$

as $(v,w)\to 0$.

With $V=W$, $Z=\mathbb R$ or $Z=\mathbb C$, and $T:V\times V\to Z$ the inner product, this gives $DT_{(v_0,w_0)}(v,w)=\langle v,w_0\rangle+\langle v_0,w\rangle$. Now if $f,g:\mathbb R\to V$ are differentiable, then $F:\mathbb R\to V\times V$ defined by $F(t)=(f(t),g(t))$ is differentiable with $DF|_t(h)=h(f'(t),g'(t))$. By the chain rule,

$$D(T\circ F)|_{t}(h) =DT|_{F(t)}\circ DF|_t(h)=h(\langle f'(t),g(t)\rangle+\langle f(t),g'(t)\rangle),$$

which means $\frac{d}{dt} \langle f, g \rangle = \langle f'(t),g(t)\rangle+\langle f(t),g'(t)\rangle$.

Since Jonas Meyer has shown above the derivative of the inner product, applying this to the $L^2$ inner product norm, we show that $$\frac{\partial}{\partial a} \langle f, f\rangle = 2\left\langle f , \frac{\partial f}{\partial a} \right\rangle$$


If $x\rightarrow \dfrac{\partial f}{\partial a}(x,a)\in L^2(X)$ , then $\int_Xf(x,a)\dfrac{\partial f}{\partial a} (x,a)dx$ makes sense. Yet, there is no reason why $D(a)=2\int_Xf(x,a)\dfrac{\partial f}{\partial a} (x,a)dx$. Indeed, this equality is associated to the derivative under the signum $\int_X$; we can do that when, for instance (Lebesgue's theorem) $|f(x,a)\dfrac{\partial f}{\partial a}(x,a)|\leq g(x)$ where $g(x)\in L^1(X)$ ($g$ does not depend on $a$).

If $x\rightarrow \dfrac{\partial f}{\partial a}(x,a)\notin L^2(X)$, then we cannot say anything about $D(a)$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.