# Integration by parts of inner product and differential

Let $$u(t),v(t)$$ be smooth curves in a vector space $$V$$ with inner product $$\langle,\rangle$$. Let $$\dot{v}(t)$$ be the differential of the curve at t, where we consider $$V$$ as a manifold. Since for a vector space manifold we can identify each tangent space with $$V$$ itself, $$\dot{v}(t)$$ is a curve in $$V$$.

I'm reading a paper where they seem to use a formula similar to

$$\int_{0}^{T} \langle u(t), \dot{v}(t) \rangle dt = - \int_{0}^{T} \langle \dot{u}(t), v(t) \rangle dt + something$$.

Could someone point me to the exact formula, and its proof?

Thank you.

• I think you're missing a $-$ sign in front of the second integral. Your paper may have a typo. – J.G. Jun 29 '19 at 12:55
• You are right, thanks, this is now corrected. – eipiplusone Jun 29 '19 at 14:28

Let $$f(t) = \langle u(t), v(t) \rangle$$. Since the inner product $$\langle \cdot, \cdot \rangle$$ is bilinear, we can use the chain rule to say $$$$\dot{f}(t) = \langle \dot{u}(t), v(t) \rangle + \langle u(t), \dot{v}(t) \rangle$$$$ (for a proof of this see Loomis and Sternberg's Advanced Calculus, Chapter $$3$$, theorem $$8.4$$) Hence, if you rearrange this equation and integrate it, we find that \begin{align} \int_0^T \langle u(t), \dot{v}(t) \rangle \, dt &= \int_0^T \left[ \dot{f}(t)- \langle \dot{u}(t), v(t) \rangle \right] \, dt \\ &= f(t) \bigg|_0^T - \int_0^T \langle \dot{u}(t), v(t) \rangle \, dt \\ &= \left[ \langle u(T), v(T) \rangle - \langle u(0), v(0) \rangle \right] - \int_0^T \langle \dot{u}(t), v(t) \rangle \, dt \end{align} Where in the second equal sign I used the Fundamental Theorem of Calculus. This is the complete statement of what you want to prove.

If you actually take a look at the statement of the theorem which I referred you to, one of the hypotheses is that the map $$\omega$$ (in conformity with their notation) is a bounded (equivalently continuous) bilinear map. So, before I invoke the use of that theorem, I actually need to justify why the inner product $$\langle \cdot, \cdot \rangle$$ is bounded. The fix is pretty easy in this case, because inner products satisfy the Cauchy Schwarz inequality, which implies boundedness. Hence, my use of the theorem is actually valid.

In general, one can prove that every multilinear transformation between finite-dimensional vector spaces is bounded (equivalently continuous), so that in particular every bilinear map between finite-dimensional vector spaces is bounded. So, if you assumed from the beginning that $$V$$ was finite-dimensional, then there is no need for further comment.

However, you did not specify the dimension of $$V$$, hence I directly argued the boundedness of the inner product by appealing to the Cauchy Schwarz inequality (which is valid even in infinite dimensions).

• @eipiplusone I added a few more words about why the use of the theorem is actually valid. Hopefully that's helpful. – peek-a-boo Jun 29 '19 at 14:51

$$\int_0^T\frac{d}{dt}\langle u(t),v(t)\rangle dt=\langle u(T),v(T)\rangle-\langle u(0), v(0)\rangle$$ by the fundamental theorem of calculus. Expanding the integrand on the left hand side and rearranging gives $$\int_0^T\langle u(t),\dot{v}(t) \rangle dt = -\int_0^T\langle \dot{u}(t),v(t)\rangle dt+\langle u(T),v(T)\rangle-\langle u(0), v(0)\rangle.$$

• Thank you. Peek-a-Boo included a reference for the derivative of the inner product (for general vector spaces), so I accepted that answer. – eipiplusone Jun 29 '19 at 14:27