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. $\endgroup$ Jul 16, 2015 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, 2015 at 19:24

2 Answers 2


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 $D(a)$.


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