Predicate<Client> hasSameNameAsOneUser =
c -> users.stream().anyMatch(u -> u.getName().equals(c.getName()));
return clients.stream()
.filter(hasSameNameAsOneUser)
.collect(Collectors.toList());
But this is quite inefficient, because it's O(m * n). You'd better create a Set of acceptable names:
Set<String> acceptableNames =
users.stream()
.map(User::getName)
.collect(Collectors.toSet());
return clients.stream()
.filter(c -> acceptableNames.contains(c.getName()))
.collect(Collectors.toList());
Also note that it's not strictly equivalent to the code you have (if it compiled), which adds the same client twice to the list if several users have the same name as the client.