Let $mu_k$, $kinmathbb N$, be a sequence of measures on a Banach space $E$ such that $(mu_k^{ast k})_{kinmathbb N}$, where $mu_k^{ast k}$ denotes the convolution, is tight, i.e. for all $varepsilon>0$ there is a compact $Ksubseteq E$ such that $$sup_{kinmathbb N}mu_k^{ast k}(K^c)<varepsilontag1.$$

Are we able to show that $(mu_k)_{kinmathbb N}$ is tight as well?

Maybe we can do something like this: Given $varepsilon$ and $K$ as above, we have $$mu_k(K^c)^k=mu_k^{bigotimes k}(times_{i=1}^kK^c)tag2,$$ where $mu_k^{bigotimes k}$ denotes the product measure, and $$mu_k^{ast k}(K^c)=theta_k(mu_k^{ast k})(K^c)tag3,$$ where $$theta_k:E^kto E;,;;;xmapstosum_{i=1}^kx_i$$ and $theta_k(mu_k^{ast k})$ denotes the pushforward measure.

Maybe we can show that $(2)$ is at most $(3)$?