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This is just a continuous extension of the discrete case, which is usually proven in an advanced calculus course. It says that given any finite sequence of non-negative real numbers x,

lim_n(Sum_i(x_i^n ))^(1/n)=max_i(x_i).

The proof in this case is simple. Indeed, we know that the limit is always greater than or equal to the max since each term in the sequence is greater or equal to the max. Thus, we only need an upper bound for each term in the sequence that converges to the max as well, and the proof will be completed via the squeeze theorem (sandwich theorem).

Set M=max_i(x_i) and k=dim(x). Since we know that each x_i is less than M, we have that the term in the limit is always less than (kM^n )^(1/n). The limit of this upper bound is easy to compute since if it exists (which it does by bounded monotonicity), then the limit must be equal to the limit of k^(1/n)M. This new limit is clearly M, since the limit of k^(1/n) is equal to 1. Since we have found an upper bound that converges to max_i(x_i), we have completed the proof.

Can you extend this proof to the continuous case?

For fun, prove the related theorem:

lim_n(Sum_i(x_i^(-n) ))^(-1/n)=min_i(x_i).