I suck at math

>>192847
Oh boy, I so much hated calculus in undergrad... I was the algebra type

OK, you sure ||.||_{C[0,1]} is the uniform norm? In my classes the uniform norm was always noted ||.||_{\infty}. Do we agree ||f||_{\infty} = sup_{x \in [a,b]} |f(x)|, for f a C^0([a,b]) function of course?

Also we agree that C^0([a,b]) is the set of continuous functions on [a,b] and C^1([a,b]) is the set of continuous functions on [a,b] that can be derived and have a continuous derivative on [a,b]?

(derivable=differentiable, I'm not sure what word is the right one in English)

The only thing I remember for sure is that you never even try to prove norm inequalities, you only prove the squared inequality, it's must easier.

>>193180
And by the way, continuously differentiable functions (i.e. class C^1 functions) are not just continuous (i.e. class C^0) but also uniformly continuous and even Lipschitz continuous and I'm pretty sure that's gonna be needed to prove that inequality, like the C constant is gonna be some sort of Lipschitz constant

Is that how you proved the case f(0)=0? Through Lipschitz continuity?

>>193198
Nah disregard what I said, the Lipschitz constant wouldn't be independent of f

By the mean value theorem of integration there is a c in [0,1] with
f(c) = \int_0^1 f'(t) dt.

Now for every x in [0,1]:

|f(x)| <= |f(c)| + |f(x) - f(c)|
= |\int_0^1 f(t) dt| + |\int_c^x f'(t) dt|
<= \int_0^1 |f(t)| dt + \int_0^1 |f'(f)| dt
<= sqrt( \int_0^1 |f(t)|^2 dt ) + sqrt( \int_0^1 |f'(f)|^2 dt )
<= sqrt(2) sqrt(\int_0^1 ( |f(t)|^2 + |f'(f)|^2 ) dt)

>>193243
>f(c) = \int_0^1 f'(t) dt.
No that's f(t) on the right-hand side

But you used it fine in your proof, now can you explain the last two inequalities? I'm not seeing any reason why they should hold true

Not OP but still interested

>>193398
They are all the Cauchy-Schwarz inequality.

>\int_0^1 |f(t)| dt
Can be intepreted as an inner product of the function |f| with the constant 1 function.
>\int_0^1 |f'(f)| dt
Can be interpreted as an inner product of the function |f'| with the constant 1 function.
>sqrt( \int_0^1 |f(t)|^2 dt ) + sqrt( \int_0^1 |f'(f)|^2 dt )
Can be interpreted as an inner product of the vector (1,1) with the vector (sqrt( \int_0^1 |f(t)|^2 dt ) , sqrt( \int_0^1 |f'(f)|^2 dt )).

>>193653
OK, didn't remember that one. That would work, indeed. Well done.

I'm a bit sad there's no simpler way to prove this result though (one that uses simpler theorems).