Gyu Eun Lee

@Evinda Instead of 2 initial conditions, think of it like this: for each $x_0$, you have a 2nd-order ODE in $t$, and you need two initial values $u(0,x_0)$ and $u_t(0,x_0)$ to solve it. Since you need 2 initial values for every $x_0$, when you vary $x_0$ you get two functions as your initial conditions.

If you like, write it as a chain of equivalences and inequalities.

What is not valid or circular about it? We have used Cauchy-Schwarz on an entirely equivalent form of the LHS to draw this conclusion.

At this point I'm not sure what you do not understand. Do you understand how to reach the conclusion of the proof from what we have worked out so far?

Think about the hint I gave you at the start...and no, that's NOT what you have.

Yes...so now what do you have in the Cauchy-Schwarz inequality?

No, that's completely off. $(a^2 + b^2) \neq (a+b)^2$. Stop doing algebra and think: how many $1/n$'s are in that sum?

Yes...which is equal to?

@LearningMath You're not taking the norm of $(1/\sqrt{n},\ldots,1/\sqrt{n})$ correctly (and you need to square it, not that it matters in this particular case)...

Can you express what I have written on the RHS as an inner product? Think of the definition of the Euclidean inner product.

But like I've said, you've got all the information you need to solve the problem, it's a matter of you actually filling in the details if you need an answer in terms of coordinates.

After this point there isn't much I can do to help you other than restate everything I've said here.

But in the case of the problem with four points $ABCD$, you need to find where $AB$ intersects this plane. Since a plane and a line can intersect at only 0, 1, or infinitely many points, and you'll get 1 intersection so long as the points are not in some weird configuration, this is enough information to find the intersection.

As to finding $b$, like I said, this alone isn't enough information in general to find $b$, it will only tell you the plane that $b$ lies in.

To determine the plane (since you are in 3-space) you need: a vector normal to the plane, and a point on the plane. The normal vector is easy: $a$ will do just fine. For the point on the plane, the projection will work.

Yes, that dotted line is the line I'm referring to. You can probably see that if you extend the dotted line and let $c$ be any vector along this line, then $c$ and $b$ have the same projections onto $a$. Now, if these points were to lie in $3$-dimensional space, then you can also revolve this dotted line to generate a plane, all of whose vectors will have the same projection onto $a$.

It means that if you know the projection of $u$ in the direction of $v$, then there is a line that you know $u$ must lie in; at least when you're in $2$ dimensions. In $3$ dimensions this is a plane. If you draw what a projection means geometrically in $2$ dimensions, then you should be able to find this line pretty easily, and in $3$ dimensions you just need to revolve this line to generate the plane. (By this I mean, go draw the picture or you won't see what I'm saying.)

$v$ is the vector you project against, it's the $a$ in your notation. But the magnitude of $v$ doesn't change the projection; only the direction $v$ is pointing in matters, so sometimes we refer to a projection in a given direction, often specified by a unit vector.