Proving bessel identity $[J_0(x)]^2 + 2[J_1(x)]^2 + 2[J_2(x)]^2 + \cdots = 1$ and $|J_0(x)|\le 1$ 
The generating function for a Bessel equation is:
$$g(x,t) = e^{(x/2)(t-1/t))}$$
Using the product $g(x,t)\cdot g(x,-t)$ show that:
a) $$[J_0(x)]^2 + 2[J_1(x)]^2 + 2[J_2(x)]^2 + \cdots = 1$$
and consequently:
b)
$$|J_0(x)|\le 1, \forall x$$
c) $$|J_n(x)| \le \frac{1}{\sqrt{2}}; n=1,2,3,\cdots$$

For a) I tried the product:
$$e^{(x/2)(t-1/t))}\cdot e^{(x/2)(-t+1/t))} = 1$$
I at least arrived at the right side of the equation. Since this generates the bessel functions, I should arrive at something related to $J_n$ in the left side. I know that
$$e^{(x/2)(t-1/t))} = \sum_{n=0\infty}^{\infty} J_n(x)t^n$$
$$e^{(x/2)(-t+1/t))} = \sum_{n=0\infty}^{\infty} J_n(x)(-t)^n$$
but it's not just a matter of multiplying coefficients from the two infinite series, right?
For $b$, I tried to use Proving Bessel equation $J_{0}(u+v) = J_0(u)\cdot J_0(v)+2\sum_{s=1}^{\infty}J_s(u)\cdot J_{-s}(v)$ but it's not as obvious
I have no idea how to deal with $c$, could somebody help me?
 A: For part (a), note that you also have
$$
g(x,-t) = g(x,1/t) = \sum_{n = 0}^\infty J_n(x) \left( \frac{1}{t} \right)^n.
$$
You can then multiply this with the original series to get
$$
\left[ \sum_{n = -\infty}^\infty J_n(x) t^{-n} \right] \left[ \sum_{n = -\infty}^\infty J_n(x) t^n \right] = 1.
$$
By multiplying these two series together and matching coefficients, you should be able to prove the given identity.
Parts (b) and (c) follow from the identity in part (a).  Note that if you have a set of quantities $a_i$ such that $a_i > 0$ for all $i$, then $a_i < \sum_j a_j$ for all $i$.
A: By the Jacobi-Anger expansion we have:
$$e^{iz\sin\theta} = \sum_{n=-\infty}^{+\infty}J_n(z) e^{in\theta}\tag{1}$$
hence by Parseval's theorem it follows that
$$ \sum_{n=-\infty}^{+\infty}\left|J_n(z)\right|^2 = \frac{1}{2\pi}\int_{0}^{2\pi}e^{iz\sin \theta} e^{iz\sin(-\theta)}\,d\theta = 1.\tag{2}$$
A: I gave some comments on the accepted answer to help clarify some doubts raised by OP. I think it is proper to avoid too much discussion in comments and hence I put all that explanation into an answer.
The generating function $g(x, t) = e^{(x/2)(t-(1/t))}$ has the property that $g(x, - t) =g(x, 1/t)$ and this means that $$\sum_{n=-\infty} ^{\infty} J_{n} (x)(-t) ^{n} =\sum_{n=-\infty} ^{\infty} J_{n} (x)(1/t)^{n}$$ or $$\sum_{n=-\infty} ^{\infty} (-1)^{n}J_{n}(x)t^{n}=\sum_{n=-\infty}^{\infty}J_{n}(x)t^{-n}=\sum_{n=-\infty}^{\infty}J_{-n}(x)t^{n}$$ where the last equality in above equation is a result of changing index of summation $n$ to $-n$ (this does not affect the sum because as $n$ takes all integer values, $-n$ also takes all integer values). Now equating coefficients of $t^{n} $ in first and last expressions of above equation we get $J_{-n}(x) =(-1)^{n}J_{n}(x)$. This relation will be needed later on. 
Next we note that $g(x, t) g(x, 1/t)=1$ and hence $$\left(\sum_{n=-\infty}^{\infty}J_{n}(x)t^{n}\right)\left(\sum_{n=-\infty}^{\infty}J_{n}(x)t^{-n}\right)=1$$ and we can express the sums by splitting positive and negative indexes to get $$\left(J_{0}(x)+\sum_{n=1}^{\infty}J_{n}(x)t^{n}+\sum_{n=1}^{\infty}J_{-n}(x)t^{-n}\right)\left(J_{0}(x)+\sum_{n=1}^{\infty}J_{-n}(x)t^{n}+\sum_{n=1}^{\infty}J_{n}(x)t^{-n}\right)=1$$ It is the next step where you are facing difficulty. Note that RHS of the above equation does not involve variable $t$, but the LHS does involve $t$. So we need to find out terms after multiplication in LHS which don't involve $t$. The way to multiply two series is multiplying each term of one series with every term of other series. The Cauchy product also does the same, but it also groups the resulting terms in a particular way to make a new series. Here we don't need that grouping. Just individual terms of the product which don't involve $t$ are needed.
Clearly one such term is obtained by multiplying first term in both series namely $J_{0}(x)J_{0}(x)=J_{0}^{2}(x)$. Next we mark the four sums involving $\sum$ notation as $A, B, C, D$ respectively. If we multiply the term with $t^{n} $ in $A$ and the term with $t^{-n} $ in $D$ then we get $J_{n} ^{2}(x)$ (the $t$'s cancel out). Similarly we multiply term with $t^{-n} $ in $B$ and term with $t^{n} $ in $C$ to get $J_{-n}^{2}(x)=(-1)^{2n}J_{n}^{2}(x)=J_{n}^{2}(x)$. Note that any other terms in the product must involve $t$. Thus the terms without $t$ in the product on LHS are $$J_{0}^{2}(x)+2J_{1}^{2}(x)+2J_{2}^{2}(x)+\cdots$$ and this is equal to $1$ by comparison with RHS. 
