# Lower bound of an expectation

Suppose a random variable $$X$$ has unit variance i.e. $$\sigma^{2} = 1$$. Is there a positive constant $$c > 0$$ such that $$\mathbb{E}[\ | X - \mathbb{E}[X] | \ ] \ge c$$

My attempt of a solution is as follows:

Let $$Y = X-\mathbb{E}(X)$$. Then by Chebyshev's inequality, for any $$k > 0$$, $$P(|Y-\mathbb{E}[Y] | \geq k) \leq 1/k^{2} \ \ (1)$$

Also by the triangle inequality,

$$P(|Y-\mathbb{E}[Y]| \geq k) \geq P(|Y| \geq k) - P(|\mathbb{E}[Y]| \geq k)$$

which implies

$$P(|Y-\mathbb{E}[Y]| \geq k) + P(|\mathbb{E}[Y]| \geq k) \geq P(|Y| \geq k)$$

or $$1/k^2 + \epsilon \geq P(|Y| \geq k)$$

where $$\epsilon \in [0, 1]$$.

From this statement, can it be concluded that $$0$$ is the greatest lower bound of $$\mathbb{E}[|Y|]$$?

I am not sure if I understood your question correctly, but for each random variable $$X$$ with positive variance, there of course is such a constant $$c(X)$$ depending on $$X$$.
For $$n \in \mathbb{N}$$ consider a random variable $$X_n$$ with $$\mathbb{P}[X_n = n] = \mathbb{P}[X_n = -n] = \frac{1}{2n^2}$$ and $$\mathbb{P}[X_n = 0] = 1 - \frac{1}{n^2}$$. Then we have $$\mathbb{E}[X_n]=0$$ and $$\mathbb{E}[X_n^2]=1$$ for all $$n$$, but also $$\mathbb{E}[\lvert X_n \rvert] = \frac{1}{n}$$. Therefore there cannot be a uniform lower bound $$c$$ without any additional assumptions.