Maximal deviation from mean of a bounded random variable

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Is there a non-probabilistic Hoeffding like inequality which tells me the deviation between a bounded random variable and its expectation?

Let $X$ be a random vector such that $||X|| \leq c$. I am looking for $$||X-\mathbb{E}[X]|| \leq ?\quad ( \text{maybe }c)$$ if such an inequality exists.

I think $\mathbb{E}[X]$ lies in the ball with radius $c/2$ and therefore $||X-\mathbb{E}[X]|| \leq c$, but I don't know how to proof it.

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$X$ lies in the ball around $0$ with radius $c$ and so does the expectation of $X$ by Jensen's inequality, since the norm is a convex function. The distance between $X$ and $\mathbb{E}[X]$ is therefore bounded almost surely by $2c$. This upper bound is tight. To see this, take your probability space to be $[0,1]$ with the uniform distribution and define the random variable $X_n$ by $$X_n(\omega)=\begin{cases} c, & \text{if }\omega\in[0,1/n) \\ -c, & \text{if }\omega\in[1/n,1] \end{cases}.$$ For each $n$ and $\omega\in[0,1/n)$, one has $$\big| X_n-\mathbb{E}[X_n]\big|=\big|c-1/n c-(1-1/n)(-c)\big|=|2(1-1/n)c|,$$ and this expression gets arbitrarily close to $2c$.