A maximal Hoeffding's inequality? Let $X_1, \cdots, X_n$ be real-valued independent random variables satisfying $|X_k|\le 1$ and $\mathbb EX_k=0$. Hoeffding's inequality tells us that for any $k=1,\cdots, n$ and $t>0$,
$$\mathbb P\Big( \Big | \frac{X_1+\cdots+ X_k}{\sqrt{k}} \Big | \ge t \Big) \le 2 e^{-t^2/2}.$$
My question is whether there exists a similar bound for the maximum over $k$. More precisely:

Question: Do there exist absolute constants $C>0$ and $A>0$ so that
  $$\mathbb P\Big( \max_{1\le k\le n} \Big | \frac{X_1+\cdots+ X_k}{\sqrt{k}} \Big | \ge t \Big) \le C e^{-t^2/A}$$
  holds for all $t>0$? If not what can we say about the left hand side?

 A: Hoeffding's inequalities for absolute values are derived by determining first the bound for the value, and then double it to arrive at a bound for the absolute value. But the question here asks for a bound related to the maximum of the absolute value, not to the absolute value of the maximum, so a direct examination of the absolute value is needed.
To compact notation, we define $S_k=\Big | X_1+\cdots+ X_k \Big |$. Then we are examining
$$\mathbb P\Big( \max_{1\le k\le n} \Big | \frac{S_k}{\sqrt{k}} \Big | \ge t\Big) = \mathbb P\Big( \bigcup_k \Big\{ \Big|\frac{S_k}{\sqrt{k}} \Big | \ge t \Big\} \Big)$$
Denote $I_Z$ the indicator function of the event $Z= \Big( \bigcup_k \Big\{ \Big|\frac{S_k}{\sqrt{k}} \Big |\ge t \Big\} \Big)$ and $I_1,..., I_n$ the indicator functions of the events, $\Big\{ \Big|\frac{S_1}{\sqrt{1}} \Big |\ge t \Big\},...\Big\{ \Big|\frac{S_n}{\sqrt{n}} \Big |\ge t \Big\} $, respectively.
Then by the properties of indicator functions
$$I_Z=\max\Big \{I_1,...,I_n\Big \}$$
Now if $I_Z =0$ it means that all $I_i,\; i=1,...,n$ are zero. If $I_Z=1$, it means that at least one of these individual indicator functions is unity, denote it $I_m$, and we have
$$I_m =1 \Rightarrow \Big|\frac{S_m}{\sqrt{m}} \Big |\ge t $$
Let $v = \text{argmax}_k \Big | \frac{S_k}{\sqrt{k}} \Big |$ and $\Big | \frac{S_v}{\sqrt{v}} \Big |$ the corresponding variable. Then in the case where $I_Z =1$ we have
$$\Big | \frac{S_v}{\sqrt{v}} \Big |\ge \Big|\frac{S_m}{\sqrt{m}} \Big |\ge t$$
Then in all cases, either when $I_Z=0$ or when $I_Z=1$ we have that, for some $h>0$,
$$I_Z \le \exp \left \{h\left (\Big | \frac{S_v}{\sqrt{v}} \Big |-t\right) \right \}$$
Therefore,
$$\mathbb P\Big( \max_{1\le k\le n} \Big | \frac{S_k}{\sqrt{k}} \Big | \ge t\Big) = \mathbb P\Big( \bigcup_k \Big\{ \Big|\frac{S_k}{\sqrt{k}} \Big | \ge t \Big\} \Big) = EI_Z \le E\exp \left \{h\left (\Big | \frac{S_v}{\sqrt{v}} \Big |-t\right) \right \}$$
$$\Rightarrow \mathbb P\Big( \max_{1\le k\le n} \Big | \frac{S_k}{\sqrt{k}} \Big | \ge t\Big)   \le e^{-ht} E\exp \left \{h \Big | \frac{S_v}{\sqrt{v}} \Big | \right \} \qquad [1] $$ 
By Hoeffding's lemma, for a random variable $Y$, with $EY=0,\;a\le Y \le b$ we have, for any real $\lambda$
$$ E\left (e^{\lambda Y} \right) \le \exp \left(\frac{\lambda ^2 (b-a)^2}{8} \right) \qquad [2]$$
In our case, we have 
$$\Big | \frac{S_v}{\sqrt{v}} \Big | = \frac{1}{\sqrt{v}}\Big |X_1 +...+ X_v\Big | \le \frac{1}{\sqrt{v}}\Big (\Big |X_1 \Big | +...+ \Big | X_v\Big | \Big) \le \frac{1}{\sqrt{v}}v = \sqrt{v}$$ 
the last inequality by the initial assumptions. Since $v\le n$ we have
$$0\le \Big | \frac{S_v}{\sqrt{v}} \Big | \le \sqrt{n} \Rightarrow -E\Big(\Big | \frac{S_v}{\sqrt{v}} \Big |\Big) \le \Big | \frac{S_v}{\sqrt{v}} \Big | - E\Big(\Big | \frac{S_v}{\sqrt{v}} \Big |\Big) \le \sqrt{n} -E\Big(\Big | \frac{S_v}{\sqrt{v}} \Big |\Big) $$.
We now have a variable with zero expected value and bounded.  The length of the interval is  $b-a = \sqrt{n} -E\Big(\Big | \frac{S_v}{\sqrt{v}} \Big |\Big) + E\Big(\Big | \frac{S_v}{\sqrt{v}} \Big |\Big) = \sqrt{n} $ and $\lambda = h $.
Inserting these values into Hoeffding's lemma and simplifying we get
$$ E\exp \Big \{h\Big | \frac{S_v}{\sqrt{v}} \Big | - hE\Big(\Big | \frac{S_v}{\sqrt{v}} \Big |\Big)\Big \} \le \exp \left(\frac{nh^2}{8} \right) $$
$$\Rightarrow E\exp \Big \{h\Big | \frac{S_v}{\sqrt{v}} \Big | \Big \} \le    \exp \left(\frac{nh^2}{8} \right)\exp\Big \{hE\Big(\Big | \frac{S_v}{\sqrt{v}} \Big |\Big)\Big \} \qquad [3]$$
Note that the exponential that moved to the right hand side does not contain any random quantities that's why the expected value has disappeared. From previously we have
$$ \Big | \frac{S_v}{\sqrt{v}} \Big | \le \sqrt{n} \Rightarrow E\Big | \frac{S_v}{\sqrt{v}} \Big | \le \sqrt{n} \Rightarrow \exp\Big \{hE\Big(\Big | \frac{S_v}{\sqrt{v}} \Big |\Big)\Big \} \le \exp\Big \{h\sqrt{n}\Big \} \le \exp\Big \{h^2n\Big \} \; [4] $$
Inserting the RHS of $[4]$ into $[3]$ and back in  $[1]$ we obtain
$$ \mathbb P\Big( \max_{1\le k\le n} \Big | \frac{S_k}{\sqrt{k}} \Big |\Big )= \mathbb P\Big( \Big | \frac{S_v}{\sqrt{v}} \Big | \ge t\Big) \le \exp \left(-ht +\frac{9nh^2}{8} \right) \qquad [5]$$
Minimizing the RHS over $h$ we obtain $h^* = \frac {4}{9n}t$ and inserting into $[5]$ we finally obtain
$$ \mathbb P\Big( \max_{1\le k\le n} \Big | \frac{S_k}{\sqrt{k}} \Big |\Big )=\mathbb P\Big( \Big | \frac{S_v}{\sqrt{v}} \Big | \ge t\Big) \le \exp\Big \{-\frac{2}{9n}t^2\Big \}  \qquad [6] $$
...which is a bound related to the number of r.v.'s involved.
A: As it stands, the maximal inequality cannot hold since we would get by letting $n$ going to infinity that
$$
\mathbb P\Big( \sup_{k\geqslant 1} \Big | \frac{X_1+\cdots+ X_k}{\sqrt{k}} \Big | \ge t \Big) \le C e^{-t^2/A}.
$$
However, by the law of the iterated logarithms, the random variable $\sup_{k\geqslant 1} \Big | \frac{X_1+\cdots+ X_k}{\sqrt{k}} \Big | $ is almost surely infinite, unless $X_i=0$ for each $i$.
However, it is possible to prove that
$$
\mathbb P\Big( \max_{1\le k\le n} \Big | \frac{X_1+\cdots+ X_k}{\sqrt{n}} \Big | \ge t \Big) \le C e^{-t^2/A}.$$
To see this, let $M_n:=\max_{1\leqslant k\leqslant n}\lvert  X_1+\cdots+ X_k\rvert$ and $S_n=\lvert  X_1+\cdots+ X_n\rvert$. By Doob's inequality,
$$
x\mathbb P\left(M_n>x\right)\leqslant \mathbb E\left[S_n\mathbf{1}_{\{M_n>x\}}\right].
$$
Write the last expectation as an integral of the tail over the positive real line and truncation this integral at $x/2$ to get that
$$
x\mathbb P\left(M_n>x\right)\leqslant 2\int_{x/2}^\infty \mathbb P\left(S_n>t\right)dt.
$$
Then use Hoeffding's inequality.
A: Look at Lemma 1 in this paper:
https://arxiv.org/pdf/1312.7308.pdf
Since you are taking a maximum the bound is actually a bit different, and the exponent is not with $t^2$ but $t^2/\log(t)$.
The idea for how to prove it is first take a naive union bound over an a geometric series $k=1,2,4,8,16$ (not necessarily with 2 as the base), then use a maximal inequality that doesn't involve the denominator of $\sqrt{k}$ (see e.g Thm 20.20 here: http://www.math.wisc.edu/~roch/grad-prob/gradprob-notes20.pdf) to bound the remaining $k$ values
