# Mean Value Inequality in Banach Space without Hahn-Banach or Integrals

If $$f : E \to F$$ is a continuous map of Banach spaces, with bounded Fréchet derivative. Then

$$x_0,x_1 \in E\Rightarrow \|f(x_1) − f(x_0)\| ≤ M\|x_1 − x_0\|$$ where $$M = \sup \|f'(x)\|.$$

The most efficient way to prove this, as far as I know, is to apply Hahn-Banach. Alternatively, one starts with $$g(t)= f(x_0 + t(x_1 − x_0));\ 0\le t\le 1$$ and reduces the problem to the real case. But, of course, the Hahn-Banach theorem is a (relatively) big gun, and to do the integral if one uses $$g$$, then integration in Banach spaces must be dealt with (via regulated functions, for example.)

The only elementary proof I have seen uses the dot product (Apostol, Rudin), but then of course, one must assume an inner product.

Below I sketch a very basic proof, but it required more effort than I expected, so my question is: can one get it cheaper, using only elementary means (very basic facts about normed spaces and the definition of derivative)?

Assume for convenience that $$x_0=f(x_0)=0$$ and consider the segment $$\{tx_1:\ 0\le t\le 1\}.$$

We have $$\|f(x)\|\le M\|x\|+\|r(x)\|\cdot\|x\|$$ where $$\frac{\|r(x)\|}{\|x\|}\to 0$$ as $$x\to 0.$$

So, if $$\epsilon>0,$$ we can choose $$\delta>0$$ such that $$t\le \min \left(1,\frac{\delta}{\|x_1\|}\right)\Rightarrow \|f(tx_1)\| ≤ (M + \epsilon)t\|x_1\|$$. To prove the claim, it will suffice to prove that $$t=1$$ satisfies this last inequality, so to that end,

let $$\tau=\sup\{r>0:\|f(tx_1)\| ≤ (M + \epsilon)t\|x_1\|\text{for all}\ t\in [0,r]\}$$, so that $$\tau\le 1$$ and $$\tau\in \{r>0:\|f(tx_1)\| ≤ (M + \epsilon)t\|x_1\|\text{for all}\ t\in [0,r]\}$$ (because $$f$$ is continuous.)

Now, toward a contradiction, suppose that $$\tau<1$$ and choose $$c>0$$ and small enough so that $$(\tau+c)x_1$$ is on the segment.

Now, there is a $$\delta' > 0$$ such that

$$\|f(x) − f(\tau x_1)\| \le (M + \epsilon)||x − \tau x_1\|<\epsilon$$ whenever $$\|x − \tau x_1\| <\delta'.$$

So, since $$\|(\tau + c)x_1 − \tau x_1\| = c\|x_1\|<\delta'$$ if $$c$$ is small enough, we have, with $$x=(\tau + c)x_1$$,

$$\|f((\tau+c) x_1)-f(\tau x_1)\| \le (M+\epsilon)\|(\tau+c)x_1-\tau x_1\|=(M+\epsilon)\|cx_1\|$$

and so finally,

$$f((\tau+c)x_1)\le (M+\epsilon)\|cx_1\|+\|f(\tau x_1)\|\le$$

$$(M+\epsilon)\|cx_1\|+ (M+\epsilon)\|\tau x_1\|=(M+\epsilon)(c+\tau)\|x_1\|$$,

which contradicts the fact that $$\tau$$ is a supremum.

• This is Theorem 1.1.1 in Hörmander's The Analysis of Partial Differential Operators. He needs 8 lines for the proof. – Jochen Mar 29 at 11:13

Here is another proof. At the core, the arguments are similar to yours (but with fewer $$\delta$$s).

Assume that there are $$x_1,y_1\in E$$ and $$\epsilon>0$$ such that $$\|f(y_1)-f(x_1) \| > (M+\epsilon) \|y_1-x_1\|.$$ By triangle inequality, one of the following inequalities are satisfied: $$\|f(y_1)-f(\frac{x_1+y_2}2) \| > (M+\epsilon) \|y_1-\frac{x_1+y_2}2\|,$$ $$\|f(\frac{x_1+y_2}2) - f(x_1)\| > (M+\epsilon) \|\frac{x_1+y_2}2-x_1\|.$$ Inductively, we can construct elements $$(x_n,y_n)$$ on the line segment $$[x_1,y_1]$$ with $$\|x_n-y_n\| = 2^{1-n} \|x_1-y_1\|$$, $$[x_{n+1},y_{n+1}]\subseteq[x_n,y_n]$$ and $$\|f(y_n)-f(x_n) \| > (M+\epsilon) \|y_n-x_n\|.$$ Then $$x_n,y_n\to x$$ for some $$x\in [x_n,y_n]$$. And $$\|f(y_n)-f(x_n) \| \le \|f(y_n)-f(x)\| + \|f(x_n)-f(x)\|.$$ By Frechet differentiability of $$f$$ at $$x$$ there we get for $$n$$ large enough $$\begin{split} \|f(y_n)-f(x_n) \| &\le \|f(y_n)-f(x)\| + \|f(x_n)-f(x)\| \\ &\le (M+\epsilon/2) (\|y_n-x\| + \|x-x_n\|) \\& = (M+\epsilon/2)\|y_n-x_n\|, \end{split}$$ which is a contradiction.

Let me try a direct proof in the flavor of your post. The proof works just with Gateaux differentiability of $$f$$ (and does not need continuity). Let $$x,h\in E$$ be given. Let $$\epsilon>0$$. Define $$I = \{ t\in [0,1]: \|f(x+th)-f(x)\| \le (M+\epsilon) t\|h\|\}.$$ Clearly, $$0\in I$$. Assume $$[0,t]\subset I$$ for some $$t\ge0$$. Then by Gateaux differentiability of $$f$$ at $$x+th$$, there is $$s_0>0$$ such that $$\|f(x+(s+t)h)-f(x+th)\| \le (M+\epsilon) s\cdot \|h\|$$ for $$s \in (0, s_0)$$. By triangle inequality, $$\|f(x+(s+t)h)-f(x)\| \le (M+\epsilon) (s+t)\cdot \|h\| \quad \forall s\in (0,s_0).$$ Hence $$[0,t+s_0)\cap [0,1]\subset I$$.

Now suppose $$[0,t)\subset I$$ for some $$t>0$$. By Gateaux differentiability of $$f$$ at $$x+th$$, there is $$s_0>0$$ such that $$\|f(x+(t-s)h)-f(x+th)\| \le (M+\epsilon) s\cdot \|h\|$$ for $$s \in (0, s_0)$$. Taking $$s\in (0,s_0$$ such that $$t-s\in I$$, and using triangle inequality as above, it follows $$t\in I$$ and $$[0,t]\subset I$$.

By induction on the reals, $$I=[0,1]$$: Assume $$I\ne[0,1]$$. Let $$t = \inf [0,1]\setminus I$$. Then $$t\ne0$$ because there is $$s_0>0$$ such that $$[0,s_0)\subset I$$. By definition of inf, $$[0,t)\subset I$$. Using the steps above: $$[0,t]\subset I$$, $$[0,t+s_0)\subset I$$ for some $$s_0>0$$.

Since $$\epsilon$$ in the definition of $$I$$ was arbitrary, the claim follows.

Using these arguments, one can prove for Gateaux differentiable $$f$$ the standard mean value inequality $$\|f(x+h)-f(x) \| \le \sup_{t\in [0,1)}\|f'(x+th)\| \|h\|,$$ with supremum taken over the half-open interval.

Can this be reduced to a supremum over the open interval $$(0,1)$$?

• You only need completeness of the (at most two-dimensional) space spanned by $x_1$ and $y_1$. – gerw Mar 29 at 7:58
• @gerw so true....... – daw Mar 29 at 8:00