Let us consider a problem where you are asked to find infimum of the set
\[\{\int_0^{1}\sqrt{1+f'(x)^2}dx\}_{f\in S}\]
where \(S\) is the set of all \(f\in C^1(\mathbb{R})\) with the property that \(f(0)=10\) and \(f(1)=0\).
When we see integral and differential together, that should remind us the famous fundamental theorem of calculus, which says that
\[\int_a^b g'(x)dx=g(b)-g(a)\].
It is unfortunate (yet interesting) that the integrand \(\sqrt{1+f'(x)^2}\) is not (immediately) of the form \(g'(x)\) for some \(g\).
If it is just \(f'(x)\) instead of \(\sqrt{1+f'(x)^2}\), we could have just applied the fundamental theorem of calculus directly.
But, that is not the case here.
Nevertheless, there is an obvious relation between \(f'(x)\) and \(\sqrt{1+(f'(x))^2}\), namely,
\[f'(x)\leq \sqrt{1+(f'(x))^2}\]
which implies that
\[\int_0^1f'(x)dx\leq \int_0^1\sqrt{1+(f'(x))^2} dx,\]
which, when applied fundamental theorem of calculus, gives us
\[f(1)-f(0)\leq \int_0^1\sqrt{1+(f'(x))^2} dx,\]
that is,
\[-10\leq \int_0^1\sqrt{1+(f'(x))^2} dx,\]
which is one of the most useless observations one can ever make.
The function \(\sqrt{1+(f'(x))^2}\) is non-negative and so the corresponding integral \(\int_0^1\sqrt{1+(f'(x))^2} dx\). So, observing
\[\int_0^1\sqrt{1+(f'(x))^2} dx\geq -10\]
is of no use for us; we already know that.
Instead of \(f(0)=10\) and \(f(1)=0\), if we had \(f(0)=0\) and \(f(1)=10\), we would have realized that
\[f(1)-f(0)\leq \int_0^1\sqrt{1+(f'(x))^2} dx,\]
which implies
\[10\leq \int_0^1\sqrt{1+(f'(x))^2} dx,\]
which may be treated as a non-trivial observation in our situation.
Instead of crying for what we do not have, we can focus on what other things can we do with what we have. It is only a matter of plus/minus.
Instead of starting with the observation \(f'(x)\leq \sqrt{1+(f'(x))^2}\), we can start with the observation \(-f'(x)\leq \sqrt{1+(f'(x))^2}\) which implies that
\[-\int_0^1f'(x)dx\leq \int_0^1\sqrt{1+(f'(x))^2} dx,\]
which, when applied fundamental theorem of calculus, gives us
\[-(f(1)-f(0))\leq \int_0^1\sqrt{1+(f'(x))^2} dx,\]
that is,
\[10\leq \int_0^1\sqrt{1+(f'(x))^2} dx,\]
which is a reasonably non-trivial observation we have made from the given data.
After patting your back for this wonderful observation, we will proceed to the next step.
As \(10\leq \int_0^1\sqrt{1+(f'(x))^2} dx\) for all \(f\in S\), we can conclude that
\[10\leq \inf\{\int_0^{1}\sqrt{1+f'(x)^2}dx\}_{f\in S}.\]
If we able to do some magic and show
\[11\leq \int_0^1\sqrt{1+(f'(x))^2} dx\]
for all \(f\in S\), we can conclude that
\[11\leq \inf\{\int_0^{1}\sqrt{1+f'(x)^2}dx\}_{f\in S},\]
which is better than saying
\[10\leq \inf\{\int_0^{1}\sqrt{1+f'(x)^2}dx\}_{f\in S}.\]
I am assuming you know why it is better? Do not prove me wrong. Give an attempt, you will figure out.
Keeping that aside, we have not computed what is \(\int_0^1\sqrt{1+(f'(x))^2} dx\) for a fixed \(f\in S\).
Forget about computing integral, do we know atleast one concrete example of a function \(f\in C^1(\mathbb{R})\) with the property that \(f(0)=10\) and \(f(1)=0\)?
It may be the case that \(S\) is an empty set and we are just looking for infimum of an empty set.
Forget about \(C^1\) function and all that. Can we find a function \(f\) with the property that \(f(0)=10\) and \(f(1)=0\); in other words, a function whose graph has the points \((0,10)\) and \((1,0)\).
High school mathematics suggest us that there is a line passing through \((0,10)\) and \((0,10)\) given by \(y=mx+c\) with \(c=10\) and \(m=-10\); that is the line \(y=10-10x\). I need not highlight this function is \(C^1\).
Let us compute \(\int_0^1\sqrt{1+(f'(x))^2} dx\) for \(f(x)=10-10x\). Note that, \(f'(x)=-10\). So,
\[\sqrt{1+(f'(x))^2}=\sqrt{101}.\]
Which implies that,
\[\int_0^1\sqrt{1+(f'(x))^2} dx=\int_0^{1}\sqrt{101}dx=\sqrt{101}\int_0^1dx=\sqrt{101}(1-0)=\sqrt{101}.\]
It is unfortunate to realize that, we do not have \(11\leq \int_0^1\sqrt{1+(f'(x))^2} dx\) for all \(f\in S\). So,
\[11> \inf\{\int_0^{1}\sqrt{1+f'(x)^2}dx\}_{f\in S}.\]
As we have got one function \(f\) with the property that \(\int_0^{10}\sqrt{1+f'(x)^2}dx=\sqrt{101}\), we see that,
\[\inf\{\int_0^{1}\sqrt{1+f'(x)^2}dx\}_{f\in S}\leq \sqrt{101}\]
Adding the previous observation
\[10\leq \inf\{\int_0^{1}\sqrt{1+f'(x)^2}dx\}_{f\in S}\]
to the current observation, we see that,
\[10\leq \inf\{\int_0^{1}\sqrt{1+f'(x)^2}dx\}_{f\in S}\leq \sqrt{101},\]
which can be seen as
\[\sqrt{100}\leq \inf\{\int_0^{10}\sqrt{1+f'(x)^2}dx\}_{f\in S}\leq \sqrt{101}\]
The gap between \(100\) and \(101\) is \(1\) and the gap between \(\sqrt{100}\) and \(\sqrt{101}\) is much less than \(1\). So, our area of search is reduced to the area between \(10=\sqrt{100}\) and \(\sqrt{101}\).
As the question is asked to you and not to me, it is not fair for me to finish it for you.
So, I will stop it here.
You may want to enjoy finishing this step.
Apologies for spoiling your fun in attempting this problem.
See you in next problem.