# pr.probability – If \$(mu_k^{ast k})\$ is tight, can we show that \$(mu_k)\$ is tight as well?

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)

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 $$(3)$$?