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This problem is quite simple however I am having some troubles to follow what exactly I am supposed to prove here.

Here is the part (a) of this problem which I have solved

For each $\lambda \ge e$ there is a unique $x \in [0,1]$ satisfying $e^x = \lambda x$

The problem is:

Consider the equation $e^x = \lambda x$

Show that the map that takes $\lambda$ to $x$ above is a continuous map on $[e, \infty)$

My question is: Am I supposed to show that the function defined by $\lambda(x) = e^x/x$ is continuous on given domain ? Or is it some other interpretation that I am missing here.

I am having unable to interpret the question correctly, any help would be appreciated.

Edit:

I got that problem actually asks for proving that map $g(\lambda) = x_1$ is continuous where $x_1$ is the unique $x$ satisfying the map $e^x =\lambda x$ . However I am still confused as to how to prove this rigorously?

Any help?

night_crawler
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  • Strange, since when $\lambda > e$ the $x$ is not unique. Do they mean the smallest solution to $e^x=\lambda x$? – Randall Feb 14 '23 at 18:41
  • @Randall, Actually the problem had two parts in first one I was asked to prove that for $x \in [0,1]$ , $e^x - \lambda x$ has a unique solution which I did, I will just update that in my question. – night_crawler Feb 14 '23 at 18:43
  • They are trying to define a function $g$ by $g(\lambda) = $ the "unique" $x$ with $e^x=\lambda x$. But, as mentioned, there is more than one $x$. – Randall Feb 14 '23 at 18:43
  • Oh, well my god, that really matters. – Randall Feb 14 '23 at 18:43
  • @Randall How should I proceed now? I am still stuck.. any hints will be helpful, I mean $g(\lambda) = x$ where $x$ is the given unique solution is obviously continuous being a constant map, is that enough to prove it? – night_crawler Feb 14 '23 at 18:52
  • that is not a constant map. How could it be? – Randall Feb 14 '23 at 19:08
  • @Randall can you explain a bit more, I am a little confused – night_crawler Feb 14 '23 at 19:18
  • Plot $h(x)= e^x-ax$ for various points $a$, and note that the smallest root moves. It's not constant. – Randall Feb 14 '23 at 19:32
  • Here: https://www.desmos.com/calculator/wqdsn2otct – Randall Feb 14 '23 at 19:34

1 Answers1

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We want to prove that the real function $g:[e,+\infty)\to\mathbb{R}$ such that $g(\lambda)\in[0,1]$ and $e^{g(\lambda)}=g(\lambda)\lambda$ for all $\lambda\geq e$ (which you have already shown to be well defined) is continuous. First of all, let $f:(0,1]\to\mathbb{R}$ be the continuous function $f(x)=e^x/x$. Then, you have probably already shown that $f$ is strictly decreasing on $(0,1]$ because $f'(x)=e^x\left(\frac{1}{x}-\frac{1}{x^2}\right)<0$ for all $x\in(0,1)$. Now, since $f$ is continuous, strictly decreasing, $f(1)=e$ and $\lim_{x\to 0^+}f(x)=+\infty$ you know that $\text{Im}(f)=[e,+\infty)$. Now the proposition that you need to use.

Proposition (Continuity of an inverse function.) If $f:I\to\mathbb{R}$ is a strictly monotonic function (not necessarily continuous) where $I$ is an interval then the function $g:f[I]\to I$ such that $g=f^{-1}$ (which exists since $f$ is injective) is continuous.

Notice that, in our case, the function $g$ of the said proposition is in fact $g:[e,+\infty)\to(0,1]$ such that $\forall\lambda\geq e:e^{g(\lambda)}=g(\lambda)\lambda$ ($g(\lambda)$ can't be $0$ since $e^x$ is never zero). Hence, we conclude that, by this proposition, $g$ is indeed continuous.

The proof of the said proposition is given by @mauro-allegranza on the linked question. However, I will also adapt it for this particular case. Let $\lambda\geq e$ and $\epsilon>0$. If $x:=g(\lambda)\in(0,1)$ (i.e. $\lambda\in(e,+\infty)$) then we know, by the fact that $f$ is strictly decreasing, that $f(x_1)>f(x)=\lambda>f(x_2)$ where $x_1\in(x-\epsilon,x)\cap(0,1]$ and $x_2\in(x,x+\epsilon)\cap(0,1]$ since $x$ is an interior point of $(0,1)$. Now choose $\delta:=\min\{f(x_1)-\lambda,\lambda-f(x_2)\}$. Then, if $\mu\in[e,+\infty)$ is such that $|\lambda-\mu|<\delta$, $$f(x_1)>\mu>f(x_2)\Rightarrow x_1< g(\mu) < x_2 \Rightarrow |g(\mu)-x|<\epsilon$$ which means that $g$ is continuous on $\lambda\in(e,+\infty)$. For the continuity of $g$ on $\lambda=e$ you can do basically the same by only making use of $x_1\in(x-\epsilon,x)\cap(0,1]$.

K. Makabre
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