# why we are looking for a bounded linear operator in definition of Frechet differentiability?

maybe this is a silly question to ask but I was wondering why in the definition of Frechet differentiability of a map between X and Y, we are saying if there exist a bounded linear operators such that makes the corresponding limit equal to zero? what is the advantage of having bounded linear operator?

• Someone else can give a better answer, but my impression of that in an infinite dimensional setting unbounded linear transformations are not nice / not simple, so it’s common to add a requirement that our linear transformation also be bounded. In finite dimensions linear transformations are guaranteed to be bounded, so this is not an issue in a finite dimensional setting. Commented May 4 at 18:14

You have to remember the whole point of differential calculus: it is to locally approximate (changes in) a given function by a linear function. For a linear map, boundedness is equivalent to Lipschitz continuity, is equivalent to continuity at one point, is equivalent to continuity at all points. Now, if your linear map was not bounded, then it is not continuous at the origin. So, in the definition $$(\Delta f)_a(h)=Df_a(h)+ o(\|h\|)$$ (where I denote $$\Delta f_a(h):=f(a+h)-f(a)$$), if we use the approximation $$(\Delta f)_a(h)\approx Df_a(h)$$, then for some values of $$h$$, the RHS would blow up. Slightly more precisely, differentiability would not imply continuity anymore (indeed an unbounded linear map $$T$$ would be everywhere differentiable according to this weakened definition, but by hypothesis, it is nowhere continuous).

More quantitatively, it gives us $$f(a+h)=f(a)+O(\|h\|)$$, which is what we get in 1-dimension (or really finite-dimensions). Without boundedness, we don’t have the big-Oh control. The class of big and little oh functions have some nice properties under compositions:

• $$O\circ O\subset O$$
• $$O\circ o\subset o$$
• $$o\circ O\subset o$$
• $$o\circ o\subset o$$.

The middle two are not true if you weaken $$O$$ to merely being continuous at the origin say. For example, take $$f(x)=\sqrt{|x|}$$ and $$g(x)=|x|^{1.2}$$. Then, $$g\in o$$ and $$f$$ is continuous at the origin, but $$(f\circ g)(x)=(g\circ f)(x)=|x|^{0.6}\notin o$$.

This has immediate consequences for chain rule (especially the middle two governing the interaction of big and little oh under composition). Supposing $$f,g$$ are functions such that all compositions are make sense, we always have $$\Delta(f\circ g)_a=(\Delta f)_{g(a)}\circ \Delta g_a$$. Suppose now $$g$$ is differentiable at $$a$$ and $$f$$ at $$g(a)$$. Then, \begin{align} \Delta(f\circ g)_a&=(\Delta f)_{g(a)}\circ \Delta g_a\\ &=[Df_{g(a)} + o]\circ [Dg_a+ o]\\ &= Df_{g(a)}\circ Dg_a+ Df_{g(a)}\circ o + o\circ Dg_a+ o\circ o. \end{align} Here, $$o$$ just denotes an arbitrary function in the class little-oh. Ok, so we have expressed the difference $$\Delta(f\circ g)_a$$ as a linear part plus three error terms. In order to get differentiability, we need to ensure that the sum of these three terms is in little-oh. Well, the last term is a composition of little-oh, so is again little-oh (by the last bullet point above). However, if you don’t have boundedness of the linear maps $$Df_{g(a)}$$ and $$Dg_a$$, then they are not bog-Oh and so you can’t conclude that these middle two terms are little-oh. So, immediately, the proof of the chain rule comes to a halt. Granted, I haven’t given you a concrete counterexample in infinite-dimensions, but I’m sure you can see how bad things can go. Also, keep in mind that the chain rule is really the first major ‘theorem’ in all of differential calculus, and without it, we can’t really do much. Perhaps we might be able to salvage some other weaker version of the chain rule, if we modify our hypotheses appropriately, but it’s not pleasant.

Another thing is that central to all of 1-differential calculus is the mean-value theorem. Similarly, in higher dimensions, the key is the mean-value inequality (where the very statement uses requires a certain supremum of operator norms be finite). Without the mean-value inequality, we don’t have Taylor’s theorem, we don’t have the conclusion that a function whose derivative is $$0$$ is constant (on connected components). Here also you might wonder whether some weaker version of the mean-value inequality exists, and perhaps it does exist. But I’m not aware of such statements (except for in 1D where I know of some generalizations of the MVT to symmetric difference quotients… but that’s not really relevant here).

In finite dimensions, people would never explicitly state boundedness as part of the hypothesis, because firstly, it is redundant (every linear map between finite-dimensional normed vector spaces is bounded), and secondly, people often focus on $$\Bbb{R}^n,\Bbb{R}^m$$, where all the estimates are done explicitly (e.g in terms of the matrix of partial derivatives). So, while it may seem like we don’t explicitly require boundedness, note that implicit in all the proofs is the use of boundedness of operators.