As long as the hash function is strong, why exactly would this be weak? Downvoting me implies you believe these people are wrong too;
https://tools.ietf.org/html/rfc6979
> It is possible to turn DSA and ECDSA into deterministic schemes by using a deterministic process for generating the "random" value k. That process must fulfill some cryptographic characteristics in order to maintain the properties of verifiability and unforgeability expected from signature schemes; namely, for whoever does not know the signature private key, the mapping from input messages to the corresponding k values must be computationally indistinguishable from what a randomly and uniformly chosen function (from the set of messages to the set of possible k values) would return.
> d. Set:
> K = HMAC_K(V || 0x00 || int2octets(x) || bits2octets(h1))
> where '||' denotes concatenation. In other words, we compute HMAC with key K, over the concatenation of the following, in order: the current value of V, a sequence of eight bits of value 0, the encoding of the (EC)DSA private key x, and the hashed message (possibly truncated and extended as specified by the bits2octets transform). The HMAC result is the new value of K. Note that the private key x is in the [1, q-1] range, hence a proper input for int2octets, yielding rlen bits of output, i.e., an integral number of octets (rlen is a multiple of 8).
