First of all, at some points (df_x)*(df_x) can be zero, so at that points

E_f(x) is not defined.

At second, if "smooth" means "everywhere differentiable", then even in the one-dimensional case of n = m = 1 the derivative may be discontinuous, so you should impose stronger restrictions on f than the property that you may expect to get for E_f(x).

@Vladimir Kadets, yes Prof. Your points are valid. I missed to mention that df_x is nonvanishing. Now I have updated the question. Thank you.

I think continuity, or even Lipschitz continuity is garanteed by a theorem of Weilandt and Hoffman concerning singular values see for example theorem 8.1.6 of (Matrix Computations -2edition, Golub-van Loan ).

As I already mentioned, even in the simplest particular case of n=m=1, the differentiability of f: R \to R does not imply continuity of its derivative f'. The standard example is the function given by formulas f(0) = 0 and f(x) = x2sin(1/x) for non-zero x. The derivative f' is everywhere defined, f'(0) = 0,

f'(x) = 2x sin(1/x) - cos(1/x) for non-zero x, and is discontinuous at point x = 0.

Vladimir Kadets I have mentioned in the description that by smooth I mean infinitely differentiable everywhere. So the function given by f(0) = 0 and f(x) = x2sin(1/x) is not smooth as its second derivative does not exist at 0.

Omran Kouba can you give more details?

RB Yadav Thank you, I did not notice this update in the description. Can you please explain what does it mean "least +ve eigen value"?

If one assumes that A(x) = (df_x)*(df_x) is invertible at all points, then the least eigenvalue is just 1/||A(x)-1||, and then you get the desired continuity by the following argument: all the entries of the matrix A(x) are continuous, the formula for the entries of the inverse matrix is a rational function on the entries of A(x) whose denominator is the determinant of A(x). Since the denominator is nowhere zero, the quotient is continuous. Finally, the norm is a continuous function on the matrix.

Vladimir Kadets thank you Prof. "least +ve eigen value " means smallest positive eigen value.

I think the case when A(x) = (df_x)*(df_x) is invertible is not very interesting. In this case f turns out to be immersion.

If you are interested in the smallest positive (i.e. non-zero) eigenvalue without requiring that the matrix is invertible, then there is no chance to get continuity. Say, it may happen that the eigenvalues of a 2x2 matrix A(x) are a(x) < b(x) that behave as follows: a(x) is positive and tends to a(z) = 0 as x goes to given point z, but b(x) tends to b(z) > 0. Then at point z the eigenvalues are 0 and b(z), so smallest positive eigenvalue is b(z). Then, E_f(x) = a(x) in the small neighborhood of z, so lim_{x \to z} E_f(x) =0, but E_f(z) = b(z), which is not equal to 0.

In other words, when the smallest eigenvalue takes value 0, the smallest POSITIVE eigenvalue jumps from being the really smallest to the second smallest one, which spoils the continuity.

On the other hand, if one speaks about the smallest eigenvalue of non-negative symmetric matrix, permitting the value 0, then it seems to me that one can demonstrate that it depends continuously on matrix.

@ Vladimir Kadets, thank you Prof. Vladimir. I am not sure whether it will help if we take f to have constant rank ( that is df_x has fixed rank for all x).

RB Yadav You have right, the condition on df_x to have a fixed rank for all x looks quite reasonable in order to get the continuity of E_f(x) .

Thank you Prof. Vladimir Kadets .

Dear RB Yadav ,

Consider the Rayleigh--Ritz quotient:

(()/()) where < , > is the Euclidean inner product

and A(x)=(df_{x})*(df_{x})

For any x, define

E(f(x)) = min_{y}{ (()/()),y≠0}

I think E(f(x)) satisfies your request.

Best wishes

Dear Prof. Issam Kaddoura , thank you for your answer. But I don't want E_f(x) to take zero values. E_f(x) is smallest positive eigen value of (df_x)*(df_x).

RB Yadav Just in order to conclude: as you suggested, consider such an f that df_x has a fixed rank (denote it k) for all x. In this case, E_f(x) is the n-k+1-th smallest singular value of the matrix df_x. So, E_f is the composition of two functions: x \to df_x and the function defined on the set of n \times m matrices which maps each matrix to its n+k-1-th singular value. Both maps are continuous, so their composition is continuous as well. The continuity of the first map is given, and for the continuity of the second one see the discussion on Stackexchange:

https://math.stackexchange.com/questions/973166/continuity-of-singular-value-map

This continuity (as already remarked Omran Kouba) can be also deduces from the inequality of Weilandt and Hoffman concerning singular values:

http://djalil.chafai.net/blog/2011/12/03/the-hoffman-wielandt-inequality/

Vladimir Kadets thanks a lot Prof. I got it. Also, thank you to all others who participated in this discussion.